Free Math Worksheets

- What key points can you read directly from intercept form? - Where is the axis of symmetry relative to the two zeros? - After finding the vertex's \(x\)-coordinate, how can you find its \(y\)-coordinate?

1. From intercept form, the zeros are \(x = 4\) and \(x = -2\). 2. The axis of symmetry is halfway between the zeros: \(x = \frac{4 + (-2)}{2} = 1\). 3. Substitute \(x = 1\): \(y = 3(1 - 4)(1 + 2) = 3 \cdot (-3) \cdot 3 = -27\). Therefore, the vertex is \((1, -27)\).

b) \((1, -27)\)

- Recall the vertex form \(y = a(x - h)^2 + k\). - Pay close attention to the sign inside the parentheses. - Which term outside the square gives the vertex's \(y\)-coordinate? - If the squared expression is just \(x^2\), what is the horizontal shift?

1. Compare each function with vertex form \(y = a(x - h)^2 + k\), whose vertex is \((h, k)\). 2. For a), \(f(x) = (x - (-3.2))^2 - 1.5\), so the vertex is \((-3.2, -1.5)\). 3. For b), \(g(x) = -(x - 5)^2 + 2\), so the vertex is \((5, 2)\). 4. For c), \(h(x) = (x - 0)^2 + 8.5\), so the vertex is \((0, 8.5)\).

a) \((-3.2, -1.5)\) b) \((5, 2)\) c) \((0, 8.5)\)

- Compare each equation with vertex form \(y=(x-h)^2+k\). - How do \(h\) and \(k\) locate the vertex? - If one of the shifts is zero, the graph is not translated in that direction.

1. For \(f(x)=x^2-5\), the \(-5\) outside the squared expression shifts the graph down \(5\) units. Its vertex is \((0, -5)\), so it matches description 2. 2. For \(g(x)=(x+4)^2\), the \(+4\) inside the squared expression shifts the graph left \(4\) units. Its vertex is \((-4, 0)\), so it matches description 1. 3. For \(h(x)=(x-1)^2+2\), the graph shifts right \(1\) unit and up \(2\) units. Its vertex is \((1, 2)\), so it matches description 3.

1. \(g(x)=(x+4)^2\), vertex \((-4, 0)\) 2. \(f(x)=x^2-5\), vertex \((0, -5)\) 3. \(h(x)=(x-1)^2+2\), vertex \((1, 2)\)

- Compare the equation with \(y=a(x-h)^2+k\). - What does the sign of \(a\) tell you? - How does \(|a|\) affect the width of the parabola? - Is the vertex a maximum or a minimum?

1. From vertex form, the vertex is \((3, 5)\). 2. The coefficient is \(a=-1.5\). Since \(a<0\), the parabola opens downward. Since \(|a|=1.5>1\), the graph is narrower than \(y=x^2\). 3. The vertex is the maximum point, so the range is \((-\infty, 5]\). 4. Starting with \(y=x^2\), reflect the graph across the x-axis, apply a vertical stretch by a factor of \(1.5\), shift it \(3\) units right, and shift it \(5\) units up.

a) \((3, 5)\) b) The parabola opens downward and is narrower than \(y=x^2\). c) \((-\infty, 5]\) d) Reflect across the x-axis, vertically stretch by a factor of \(1.5\), shift \(3\) units right, and shift \(5\) units up.

- Expand the squared binomial before distributing the coefficient. - Set the function equal to \(0\) to find the ground-level intercepts. - The width is the distance between the two zeros.

1. Expand the square: \((x - 10)^2 = x^2 - 20x + 100\). 2. Substitute and simplify: \(p(x) = -0.1(x^2 - 20x + 100) + 10 = -0.1x^2 + 2x - 10 + 10\). Therefore, \(p(x) = -0.1x^2 + 2x\). 3. Set the standard form equal to zero: \(-0.1x^2 + 2x = 0\). Factoring gives \(x(-0.1x + 2) = 0\), so the zeros are \(x = 0\) and \(x = 20\). 4. The distance between the ground-level intercepts is \(20 - 0 = 20\), so the arch is \(20\,\text{m}\) wide.

a) \(p(x) = -0.1x^2 + 2x\) b) The zeros are \(x = 0\) and \(x = 20\), so the width is \(20\,\text{m}\).

- What information can you read directly from vertex form? - How does the sign inside the parentheses determine the vertex's \(x\)-coordinate? - What equation do you solve to find \(x\)-intercepts? - How do you solve an equation of the form \((x - h)^2 = r\)?

1. For a), vertex form shows the vertex is \((-4, -9)\). Set \((x + 4)^2 - 9 = 0\), so \((x + 4)^2 = 9\). Therefore, \(x + 4 = \pm 3\), giving \(x = -7\) or \(x = -1\). 2. For b), vertex form shows the vertex is \((3, 8)\). Set \(-2(x - 3)^2 + 8 = 0\), so \((x - 3)^2 = 4\). Therefore, \(x - 3 = \pm 2\), giving \(x = 1\) or \(x = 5\).

a) Vertex \((-4, -9)\); zeros \(x = -7\) and \(x = -1\) b) Vertex \((3, 8)\); zeros \(x = 1\) and \(x = 5\)

- Locate the vertex relative to the x-axis. - Identify whether the parabola opens upward or downward. - Use those two features to decide whether the graph crosses, touches, or misses the x-axis.

1. For a), the vertex is \((2, 12)\), which is above the x-axis. Because the parabola opens downward, it crosses the x-axis twice. 2. For b), the vertex is above the x-axis and the parabola opens upward, so the entire graph stays above the x-axis. It has no real zeros. 3. For c), the vertex is \((-3, 0)\), which lies on the x-axis. The parabola touches the x-axis once, so it has one real zero.

a) Two real zeros b) No real zeros c) One real zero

- What form does a quadratic have when its vertex is at the origin? - Substitute the given point to find the leading coefficient. - Set the function equal to the requested height and solve for the distance.

1. Because the vertex is at the origin, use \(f(x) = ax^2\). 2. Substitute the point \((4.00, 3.20)\): \(3.20 = a \cdot 4.00^2 = 16a\), so \(a = 0.2\). Therefore, \(f(x) = 0.2x^2\). 3. Set the height equal to \(2.00\): \(2.00 = 0.2x^2\). Then \(x^2 = 10\), so the two coordinates are \(x = \pm\sqrt{10}\). 4. The requested horizontal distance from the center is positive: \(\sqrt{10}\,\text{m} \approx 3.16\,\text{m}\).

a) \(f(x) = 0.2x^2\) b) \(\sqrt{10}\,\text{m} \approx 3.16\,\text{m}\) from the center

- Set the function equal to zero to find its zeros. - Look for a perfect-square trinomial after clearing the decimal coefficient. - The exponent on a linear factor gives the multiplicity of its zero. - Decide whether an even-multiplicity zero makes the graph cross or only touch the x-axis.

1. Set the function equal to zero: \(0.5x^2 - 4x + 8 = 0\). Multiplying both sides by \(2\) gives \(x^2 - 8x + 16 = 0\). 2. Factor the perfect-square trinomial: \((x - 4)^2 = 0\). Therefore, the only zero is \(x = 4\). 3. Including the original leading coefficient, the factored form is \(f(x) = 0.5(x - 4)^2\). 4. The factor \(x - 4\) occurs twice, so the zero has multiplicity 2. The graph touches the x-axis at \((4, 0)\) and does not cross it.

a) \(x = 4\) b) \(f(x) = 0.5(x - 4)^2\) c) The zero has multiplicity 2. The graph touches the x-axis at \((4, 0)\) without crossing it.

- In vertex form, \(k\) is the y-coordinate of the vertex. - For an upward-opening parabola, the vertex gives the minimum value. - A downward-opening parabola with a positive maximum crosses the x-axis twice.

1. The vertex of \(h\) is \((5, k)\), and the parabola opens upward. 2. An upward-opening parabola has no real zeros when its minimum value is positive. Therefore, \(h\) has no real zeros when \(k>0\). 3. A downward-opening parabola has two real zeros when its vertex lies above the x-axis. For example, \(p(x)=-x^2+1\) opens downward and has zeros \(x=-1\) and \(x=1\).

a) \(k>0\) b) One example is \(p(x)=-x^2+1\).

- Identify the vertex of each parabola. - In \(y = (x - d)^2 + e\), the vertex is \((d, e)\). - Watch the sign inside the parentheses.

1. Graph \(f\) has vertex \((-2, 0)\), so it matches (1), \(y = (x + 2)^2\). 2. Graph \(g\) has vertex \((1.5, 0)\), so it matches (2), \(y = (x - 1.5)^2\). 3. Graph \(h\) has vertex \((0, -3)\), so it matches (3), \(y = x^2 - 3\). 4. Graph \(k\) has vertex \((4, 1)\), so it matches (4), \(y = (x - 4)^2 + 1\). 5. Equation (5) would have vertex \((-3, 0)\), which is not shown.

\(f \rightarrow\) (1) \(g \rightarrow\) (2) \(h \rightarrow\) (3) \(k \rightarrow\) (4) Equation (5) is unused.

- Identify the vertex of each parabola. - Use a point one unit from the vertex to determine the vertical scale factor. - The x-intercepts occur where the graph crosses the x-axis.

1. The vertex of \(f\) is \((2, -1)\). The parabola has the standard width and opens upward, so \(f(x) = (x - 2)^2 - 1\). 2. The vertex of \(g\) is \((-1, 2)\). Moving one unit right from the vertex lowers the graph by \(0.5\), so \(a = -0.5\). Thus, \(g(x) = -0.5(x + 1)^2 + 2\). 3. From the graph, the x-intercepts of \(f\) are approximately \(1\) and \(3\). The x-intercepts of \(g\) are approximately \(-3\) and \(1\).

a) \(f(x) = (x - 2)^2 - 1\); \(g(x) = -0.5(x + 1)^2 + 2\) b) \(f\): \(x \approx 1\) and \(x \approx 3\); \(g\): \(x \approx -3\) and \(x \approx 1\)

- Move the constant so one side is \(0\). - Zeros are the x-coordinates where the graph crosses the x-axis. - Substitute each value into the original equation.

1. Subtract \(3\) from both sides: \(x^2 - 2x - 3 = 0\). 2. The graph crosses the x-axis at \(x = -1\) and \(x = 3\). 3. For \(x = -1\), \((-1)^2 - 2(-1) = 3\). 4. For \(x = 3\), \(3^2 - 2(3) = 3\).

a) \(x^2 - 2x - 3 = 0\) b) \(x = -1\) and \(x = 3\) c) Both values satisfy the original equation.

- Compare the opening direction and a clear point on the graph with each option. - Find the zeros to locate the axis of symmetry. - The positive zero represents where the stream returns to the water surface.

1. The graph opens downward, so function (3) cannot match. It also passes through \((2, 3)\). For function (1), \(f(2) = -0.25 \cdot 2^2 + 2 \cdot 2 = 3\). Function (2) gives \(g(2) = 6\). Therefore, function (1) matches the graph. 2. Factor \(f(x) = -0.25x(x - 8)\). Its zeros are \(x = 0\) and \(x = 8\), so the vertex lies midway at \(x = 4\). 3. Evaluate \(f(4) = -0.25 \cdot 4^2 + 2 \cdot 4 = 4\). The maximum height is \(4\,\text{m}\), matching the graph. 4. The positive zero \(x = 8\) represents the point where the stream returns to the water surface, so the horizontal range is \(8\,\text{m}\).

a) Function (1), \(f(x) = -0.25x^2 + 2x\) b) \(4\,\text{m}\) at \(x = 4\) c) \(8\,\text{m}\)

- Read the graph's height at the requested horizontal distance. - The highest point is the vertex. - The landing point is the positive x-intercept.

1. At \(x = 1\), the graph has height \(y = 3\). The ball is \(3\,\text{ft}\) high. 2. The vertex is \((2, 4)\), so the maximum height is \(4\,\text{ft}\) after a horizontal distance of \(2\,\text{ft}\). 3. The path returns to the x-axis at \((4, 0)\), so the ball lands after \(4\,\text{ft}\).

1) \(3\,\text{ft}\) 2) A maximum height of \(4\,\text{ft}\) after \(2\,\text{ft}\) horizontally 3) \(4\,\text{ft}\)

- What information can you read directly from vertex form? - What x-value is used to find a y-intercept? - Consider the direction the parabola opens and the location of its vertex relative to the x-axis. - What equation must be true when an x-value is a zero of a function?

1. The equation is in vertex form \(a(x-h)^2+k\). Here, \(h=-4\) and \(k=-16\), so the vertex is \((-4, -16)\). 2. For the y-intercept, substitute \(x=0\): \(g(0)=(0+4)^2-16=16-16=0\). The y-intercept is \((0, 0)\). 3. The parabola opens upward because \(a=1>0\), and its vertex is below the x-axis. Therefore, the graph crosses the x-axis twice, so \(g\) has two real zeros. 4. Substitute \(x=2\): \(g(2)=(2+4)^2-16=36-16=20\). Since \(g(2)\ne 0\), \(x=2\) is not a zero.

a) \((-4, -16)\) b) \((0, 0)\) c) Two real zeros; the parabola opens upward and its vertex is below the x-axis. d) No. Since \(g(2)=20\ne 0\), \(x=2\) is not a zero.

- How does the sign of the leading coefficient affect the direction the parabola opens? - Use the vertex to identify the greatest or least possible output. - For each point, compare its y-coordinate with the function value at the same x-coordinate.

1. The vertex is \((1, 3)\). Because the leading coefficient \(-2\) is negative, the parabola opens downward, so its greatest y-value is \(3\). The range is \((-\infty, 3]\). 2. At \(x=0\), \(f(0)=-2(0-1)^2+3=1\). Point \(A\) has y-coordinate \(2\), and \(2>1\), so \(A\) lies above the graph. 3. At \(x=2\), \(f(2)=-2(2-1)^2+3=1\). Point \(B\) has y-coordinate \(0\), and \(0<1\), so \(B\) lies below the graph.

a) \((-\infty, 3]\) b) \(A(0, 2)\) lies above the graph because \(f(0)=1<2\). \(B(2, 0)\) lies below the graph because \(f(2)=1>0\).

- Use the linear equation and the factored quadratic form to find zeros. - A shared zero gives a common point on the x-axis. - Compare function values at the same x-coordinate.

1. For \(h(x) = 0\), \(2x + 4 = 0\), so \(x = -2\). 2. From the factored form of \(k\), its zeros are \(x = -2\) and \(x = 3\). 3. Both functions have value \(0\) at \(x = -2\), so the graphs share the point \((-2, 0)\) on the x-axis. 4. Since \(h(1) = 2 \cdot 1 + 4 = 6\), \(S = (1, 6)\). Also, \(k(1) = 3 \cdot (-2) = -6\), so \(S\) lies above the graph of \(k\). 5. At \(x = 0\), \(h(0) = 4\) and \(k(0) = 2 \cdot (-3) = -6\). The vertical distance is \(|4 - (-6)| = 10\).

a) \(h\): \(x = -2\); \(k\): \(x = -2\) and \(x = 3\) b) The graphs share \((-2, 0)\). c) \(y_S = 6\), and \(S\) lies above the graph of \(k\). d) \(10\) units

- Which parts of vertex form are determined directly by the vertex? - How can the given point be used to determine \(a\)? - What happens to every term when you distribute a negative sign? - How can you expand the vertex form to obtain standard form?

1. Substitute the vertex into \(y=a(x-h)^2+k\): \(y=a(x+2)^2+5\). 2. Use the point \((0, 1)\): \(1=a(0+2)^2+5\). 3. Solve for \(a\): \(1=4a+5\), so \(-4=4a\) and \(a=-1\). 4. The vertex form is \(y=-(x+2)^2+5\). 5. Expand the squared binomial: \(y=-(x^2+4x+4)+5\). 6. Combine like terms: \(y=-x^2-4x+1\).

The vertex form is \(y=-(x+2)^2+5\). The standard form is \(y=-x^2-4x+1\).

- Which form is useful when the zeros are known? - How can the additional point determine the leading coefficient? - Where is the axis of symmetry relative to the two zeros? - How do you expand intercept form into standard form?

1. Write the function in intercept form: \(f(x) = a(x + 2)(x - 4)\). 2. Substitute \(P(0, -4)\): \(-4 = a \cdot 2 \cdot (-4) = -8a\), so \(a = 0.5\). 3. Expand: \(f(x) = 0.5(x + 2)(x - 4) = 0.5x^2 - x - 4\). 4. The axis of symmetry is the midpoint of the zeros: \(x = \frac{-2 + 4}{2} = 1\). 5. Evaluate \(f(1) = 0.5 \cdot 3 \cdot (-3) = -4.5\). Therefore, the vertex is \((1, -4.5)\).

\(f(x) = 0.5x^2 - x - 4\); vertex \((1, -4.5)\)

- What information can you read directly from vertex form? - What does the sign of the coefficient outside the squared expression tell you? - What is the y-value at an x-intercept? - How does a coefficient between \(0\) and \(1\) affect the width of a parabola?

1. The equation is in vertex form \(a(x-h)^2+k\). Here, \(h=-1\) and \(k=-2\), so the vertex is \((-1, -2)\). Since \(a=\frac{1}{2}>0\), the parabola opens upward. 2. Set \(f(x)=0\): \(\frac{1}{2}(x+1)^2-2=0\). 3. Add \(2\) and multiply by \(2\): \((x+1)^2=4\). 4. Take both square roots: \(x+1=2\) or \(x+1=-2\). 5. Therefore, the zeros are \(x=1\) and \(x=-3\). 6. The vertex is \((-1, -2)\), the x-intercepts are \((-3, 0)\) and \((1, 0)\), and the graph is symmetric about \(x=-1\). Since \(0<\frac{1}{2}<1\), the upward-opening parabola is wider than \(y=x^2\).

1. The vertex is \((-1, -2)\), and the parabola opens upward. 2. The zeros are \(x=-3\) and \(x=1\). 3. The graph is an upward-opening parabola with vertex \((-1, -2)\), axis of symmetry \(x=-1\), and x-intercepts \((-3, 0)\) and \((1, 0)\). It is wider than \(y=x^2\).

- How are the vertex coordinates represented in vertex form? - Use the square of a binomial to expand the expression. - Is the vertex included in the restricted domain? - Compare the function values at the vertex and the endpoints of the interval.

1. A translation of \(y=x^2\) has leading coefficient \(1\). Using the vertex \((3, -2)\), the vertex form is \(f(x)=(x-3)^2-2\). 2. Expand: \(f(x)=x^2-6x+9-2=x^2-6x+7\). 3. The vertex x-coordinate \(3\) lies in \([1, 5]\). Since the parabola opens upward, the minimum value on the interval is \(-2\). 4. Evaluate the endpoints: \(f(1)=(1-3)^2-2=2\) and \(f(5)=(5-3)^2-2=2\). 5. Therefore, the maximum value on the interval is \(2\), and the range is \([-2, 2]\).

a) Vertex form: \(f(x)=(x-3)^2-2\) Standard form: \(f(x)=x^2-6x+7\) b) \([-2, 2]\)

- Set the two function expressions equal to find common points. - What does the number of solutions of the resulting equation tell you about the graphs? - How does the sign outside the squared expression affect the direction the parabola opens? - How does the y-coordinate of the vertex determine the range?

1. At an intersection, the function values are equal: \(x^2=-(x-2)^2+2\). 2. Expand and simplify: \(x^2=-x^2+4x-2\), so \(2x^2-4x+2=0\). 3. Divide by \(2\) and factor: \(x^2-2x+1=0\), so \((x-1)^2=0\). 4. The only solution is \(x=1\). Then \(y=f(1)=1\), so the graphs have one intersection point, \((1, 1)\). 5. The vertex of \(g\) is \((2, 2)\). Because the coefficient of the squared expression is negative, the parabola opens downward. 6. The vertex gives the maximum value \(2\), so the range is \((-\infty, 2]\).

a) The graphs have exactly one intersection point: \((1, 1)\). b) \((-\infty, 2]\), because \(g\) opens downward and has a maximum value of \(2\) at its vertex.

- Substitute each point's x- and y-coordinates into the equation. - Rearrange the equation so that the parameter is isolated. - Calculate the parameter for each point, then identify repeated values.

1. For a point \((x, y)\) on the graph, solve \(y=x^2+k\) for the parameter: \(k=y-x^2\). 2. For \(A(2, 1.5)\), \(k=1.5-2^2=-2.5\). 3. For \(B(-1, -1.5)\), \(k=-1.5-(-1)^2=-2.5\). 4. For \(C(0, 4)\), \(k=4-0^2=4\). 5. For \(D(3, 13)\), \(k=13-3^2=4\). 6. For \(E(1, -2)\), \(k=-2-1^2=-3\). 7. The three distinct values are \(-2.5\), \(4\), and \(-3\).

The three values of \(k\) are \(-2.5\), \(4\), and \(-3\).

- What form does a vertical translation of \(y=x^2\) have? - What does it mean for the vertex to lie on the y-axis? - Substitute the coordinates of the given point into the equation. - What function value corresponds to a zero?

1. A translation of \(y=x^2\) with its vertex on the y-axis has the form \(f(x)=x^2+k\). 2. Substitute \(P(3, 5)\): \(5=3^2+k\), so \(5=9+k\) and \(k=-4\). Therefore, \(f(x)=x^2-4\). 3. To find the zeros, set \(f(x)=0\): \(x^2-4=0\). 4. Then \(x^2=4\), so \(x=-2\) or \(x=2\).

1. \(f(x)=x^2-4\) 2. The zeros are \(x=-2\) and \(x=2\).

- Rewrite each equation in vertex form. - Locate each vertex relative to the x-axis and note the direction the parabola opens. - Decide whether each graph crosses, touches, or does not reach the x-axis.

1. Rewrite \(p_1\): \(x^2-4x+4=(x-2)^2\). Its vertex is \((2, 0)\). 2. The first parabola opens upward and its vertex lies on the x-axis, so it touches the x-axis once. Thus, \(p_1\) has one distinct real zero. 3. Rewrite \(p_2\): \(x^2-4x+5=(x-2)^2+1\). Its vertex is \((2, 1)\). 4. The second parabola opens upward and its minimum value is \(1\), so it never reaches the x-axis. Thus, \(p_2\) has no real zeros.

For \(p_1\), the vertex is \((2, 0)\), and the function has one distinct real zero. For \(p_2\), the vertex is \((2, 1)\), and the function has no real zeros.

- Identify the vertex from the function’s form. - Relate the vertex’s vertical position to the number of x-intercepts. - For part b), substitute one of the given zeros into the function.

1. The function is in vertex form, with vertex \((3, k)\), and the parabola opens upward. 2. For exactly one real solution, the vertex must lie on the x-axis, so \(k = 0\). 3. For the solutions \(1\) and \(5\), substitute either value. Using \(x = 1\), \((1 - 3)^2 + k = 0\), so \(4 + k = 0\) and \(k = -4\). Substituting \(x = 5\) gives the same value. 4. Because \((x - 3)^2 \ge 0\), the expression \((x - 3)^2 + k\) is always positive when \(k > 0\). Therefore, there are no real solutions for \(k > 0\).

a) \(k = 0\) b) \(k = -4\) c) \(k > 0\)

- Rewrite each function as the square of a binomial. - Locate the two vertices on the x-axis. - Find the horizontal distance and direction from the vertex of \(f\) to the vertex of \(g\).

1. Rewrite each perfect-square trinomial: \(f(x)=(x-2)^2\) and \(g(x)=(x+5)^2\). 2. Therefore, the vertices are \((2, 0)\) for \(f\) and \((-5, 0)\) for \(g\). 3. The horizontal change from \(2\) to \(-5\) is \(-5-2=-7\). 4. Translate the graph of \(f\) \(7\) units left to make it coincide with the graph of \(g\).

a) The vertex of \(f\) is \((2, 0)\), and the vertex of \(g\) is \((-5, 0)\). b) Translate the graph of \(f\) \(7\) units left.

- What does it tell you when two points on a parabola have the same \(y\)-coordinate? - Which coordinate of the vertex equals the minimum value? - How do you write vertex form once you know the vertex?

1. The points \((-2, 11)\) and \((4, 11)\) have the same \(y\)-coordinate, so they are symmetric about the axis of symmetry. The axis is halfway between their \(x\)-coordinates: \(x = \frac{-2 + 4}{2} = 1\). 2. The minimum value is the \(y\)-coordinate of the vertex, so \(k = 2\). 3. The vertex is \((1, 2)\), and the function is \(f(x) = (x - 1)^2 + 2\).

The vertex is \((1, 2)\), and the function is \(f(x) = (x - 1)^2 + 2\).

- Where is the axis of symmetry when two points on a parabola have the same \(y\)-coordinate? - Once you know the vertex's \(x\)-coordinate, how can you start writing vertex form? - How do you expand vertex form and compare coefficients?

1. The points have the same \(y\)-coordinate, so the axis of symmetry is halfway between their \(x\)-coordinates: \(x = \frac{1 + 5}{2} = 3\). Therefore, the vertex has \(x\)-coordinate \(3\). 2. Because the leading coefficient is \(1\), write \(g(x) = (x - 3)^2 + k\). Substitute \((5, 5)\): \(5 = (5 - 3)^2 + k\), so \(k = 1\). 3. Expand: \(g(x) = (x - 3)^2 + 1 = x^2 - 6x + 10\). 4. Comparing coefficients with \(x^2 + px + q\) gives \(p = -6\) and \(q = 10\).

a) The vertex has \(x\)-coordinate \(3\), the midpoint of \(1\) and \(5\). b) \(p = -6\) and \(q = 10\)

- What are the \(y\)-coordinates of the zeros? - How are the zeros related to the axis of symmetry? - What does a leading coefficient of \(-1\) tell you about the form and direction of the parabola?

1. The axis of symmetry is halfway between the zeros: \(x = \frac{-3 + 5}{2} = 1\). 2. Since the leading coefficient is \(-1\), write \(f(x) = -(x - 1)^2 + k\). 3. Substitute the zero \((5, 0)\): \(0 = -(5 - 1)^2 + k = -16 + k\), so \(k = 16\). 4. Therefore, \(f(x) = -(x - 1)^2 + 16\), the vertex is \((1, 16)\), and the maximum value is \(16\).

The vertex is \((1, 16)\), the function is \(f(x) = -(x - 1)^2 + 16\), and the maximum value is \(16\).

- Can you rewrite the quadratic in vertex form by completing the square? - Can the quadratic be factored to find its zeros? - How does a horizontal shift change the value inside the squared binomial?

1. Complete the square: \(f(x) = x^2 + 6x + 9 - 9 + 5 = (x + 3)^2 - 4\). Therefore, the vertex is \((-3, -4)\). 2. Factor the original expression: \(x^2 + 6x + 5 = (x + 1)(x + 5)\). Thus, the zeros are \(x = -1\) and \(x = -5\). 3. Shifting the graph \(3\) units left moves the vertex from \((-3, -4)\) to \((-6, -4)\). Therefore, \(g(x) = (x + 6)^2 - 4\).

a) \((-3, -4)\) b) \(x = -5\) and \(x = -1\) c) \(g(x) = (x + 6)^2 - 4\)

- How does factoring the leading coefficient help you complete the square? - Can you solve for the zeros directly from vertex form? - What happens to a point's coordinates when the graph is reflected across the \(x\)-axis?

1. Factor \(-0.5\) from the variable terms and complete the square: \(h(x) = -0.5(x^2 - 4x) + 6 = -0.5((x - 2)^2 - 4) + 6 = -0.5(x - 2)^2 + 8\). The vertex is \((2, 8)\). 2. Set the vertex form equal to zero: \(-0.5(x - 2)^2 + 8 = 0\). Then \((x - 2)^2 = 16\), so \(x - 2 = \pm 4\). The zeros are \(x = -2\) and \(x = 6\). 3. Reflecting across the \(x\)-axis changes the vertex to \((2, -8)\). Shifting up \(2\) units changes it to \((2, -6)\).

a) \(h(x) = -0.5(x - 2)^2 + 8\); vertex \((2, 8)\) b) \(x = -2\) and \(x = 6\) c) \((2, -6)\)

- Begin with vertex form using the given vertex. - Which values of \(a\) produce an upward-opening parabola wider than \(y=x^2\)? - Use the vertex and opening direction to determine the range. - What changes when only the sign of \(a\) is reversed?

1. Using the vertex, write \(g(x)=a(x+2)^2-4\). 2. Opening upward requires \(a>0\), and being wider than \(y=x^2\) requires \(0<a<1\). One possible choice is \(a=0.5\), giving \(g(x)=0.5(x+2)^2-4\). 3. This parabola opens upward and has minimum value \(-4\), so its range is \([-4,\infty)\). 4. Multiplying \(a\) by \(-1\) makes the parabola open downward without changing the vertex. The vertex becomes the maximum, so the new range is \((-\infty,-4]\).

a) One possible equation is \(g(x)=0.5(x+2)^2-4\). Any value of \(a\) with \(0<a<1\) works. b) \([-4,\infty)\) c) The range becomes \((-\infty,-4]\) because the parabola opens downward and has maximum value \(-4\).

- Which parts of the two vertex-form equations are the same? - Where does an upward-opening parabola change from decreasing to increasing? - Substitute the point's x-coordinate into each function. - Compare each function value with the point's y-coordinate.

1. Both equations have vertex form \(a(x-h)^2+k\) with \(h=-2\) and \(k=-1\). Thus, both graphs have vertex \((-2, -1)\) and axis of symmetry \(x=-2\). 2. Both leading coefficients are positive, so both parabolas open upward. An upward-opening parabola is strictly decreasing to the left of and through its vertex. Therefore, both functions are strictly decreasing on \((-\infty,-2]\). 3. Evaluate \(f(0)\): \(f(0)=0.5(0+2)^2-1=0.5\cdot 4-1=1\). Thus, \(P\) lies on the graph of \(f\). 4. Evaluate \(h(0)\): \(h(0)=3(0+2)^2-1=3\cdot 4-1=11\). Since \(11\ne 1\), \(P\) does not lie on the graph of \(h\).

a) Both graphs have vertex \((-2, -1)\) and axis of symmetry \(x=-2\). b) \((-\infty,-2]\) c) The point \(P(0, 1)\) lies only on the graph of \(f\).

- What can you read directly from intercept form? - Where is the axis of symmetry relative to the two zeros? - Once you know the vertex, how do you write vertex form with the same leading coefficient?

1. The zeros are \(x = -1\) and \(x = 7\), so the axis of symmetry is their midpoint: \(x = \frac{-1 + 7}{2} = 3\). 2. Evaluate the function at \(x = 3\): \(h(3) = 0.25(3 + 1)(3 - 7) = 0.25 \cdot 4 \cdot (-4) = -4\). The vertex is \((3, -4)\). 3. Keep the leading coefficient \(0.25\) and use the vertex: \(h(x) = 0.25(x - 3)^2 - 4\).

\(h(x) = 0.25(x - 3)^2 - 4\)

- How can you rewrite the quadratic in vertex form? - What is the \(y\)-coordinate of any point on the \(x\)-axis? - How does a vertical shift change the vertex?

1. Complete the square: \(y = x^2 - 10x + 25 - 25 + 21 = (x - 5)^2 - 4\). 2. The vertex is \((5, -4)\). 3. To move the vertex from \((5, -4)\) to the \(x\)-axis, shift the graph \(4\) units up.

1) \((5, -4)\) 2) Shift the graph \(4\) units up.

- Can you factor the equation used to find the \(x\)-intercepts? - Which coefficient controls the opening direction and width of a parabola? - What does vertex form look like when the vertex lies on the \(y\)-axis?

1. Set \(h(x) = 0\): \(-x^2 + 4x - 3 = 0\), or \(x^2 - 4x + 3 = 0\). Factoring gives \((x - 1)(x - 3) = 0\), so the \(x\)-intercepts are \((1, 0)\) and \((3, 0)\). 2. The original leading coefficient is \(-1\), so the translated graph must keep leading coefficient \(-1\). 3. A quadratic with vertex \((0, 2)\) and leading coefficient \(-1\) is \(g(x) = -(x - 0)^2 + 2 = -x^2 + 2\).

a) \((1, 0)\) and \((3, 0)\) b) \(g(x) = -x^2 + 2\)

- Translate the verbal description into a product involving one variable. - Expand the product to obtain a quadratic. - Which feature of an upward-opening parabola gives its minimum?

1. Let \(x\) be the number. The product is \(P(x) = 3x(x - 5) = 3x^2 - 15x\). 2. Since the leading coefficient is positive, the quadratic has a minimum at its vertex. The vertex occurs at \(x = -\frac{-15}{2 \cdot 3} = 2.5\). 3. Evaluate the product: \(P(2.5) = 3 \cdot 2.5 \cdot (2.5 - 5) = -18.75\).

The number is \(2.5\), and the minimum product is \(-18.75\).

- Express the second number in terms of the first. - Write the product as a quadratic function. - Use the vertex to find the minimum.

1. Let the first number be \(x\). Then the second number is \(x - 10\). 2. Their product is \(P(x) = x(x - 10) = x^2 - 10x\). 3. The parabola opens upward, so its minimum occurs at the vertex: \(x = -\frac{-10}{2} = 5\). 4. The second number is \(5 - 10 = -5\), and the product is \(5 \cdot (-5) = -25\).

The numbers are \(5\) and \(-5\), and the minimum product is \(-25\).

- Write the function in vertex form. - Compare the height of the horizontal line with the minimum value. - Use the opening direction of the parabola.

1. Complete the square: \(f(x) = x^2 - 4x + 4 + 3 = (x - 2)^2 + 3\). The vertex is \((2, 3)\). 2. The parabola opens upward and has minimum value \(3\). 3. A horizontal line intersects it twice exactly when the line lies above the vertex, so \(c > 3\). 4. When \(c = 2\), the line lies below the minimum value, so there are no intersections.

a) \((2, 3)\) b) \(c > 3\) c) No intersections

- Express one number in terms of the other using their sum. - Write the sum of squares as a quadratic function. - Use the vertex to find the minimum.

1. Let the numbers be \(x\) and \(y\). Since \(x + y = 20\), \(y = 20 - x\). 2. The sum of their squares is \(S(x) = x^2 + (20 - x)^2 = 2x^2 - 40x + 400\). 3. This quadratic opens upward, so its minimum occurs at the vertex: \(x = -\frac{-40}{2 \cdot 2} = 10\). 4. Then \(y = 20 - 10 = 10\), and the minimum sum of squares is \(10^2 + 10^2 = 200\).

The numbers are \(10\) and \(10\), and the minimum sum of their squares is \(200\).

- Which form uses the zeros directly? - How can the additional point determine the leading coefficient? - How do you expand a product into standard form?

1. Use intercept form: \(f(x) = a(x + 1)(x - 3)\). 2. Substitute \(P(1, -4)\): \(-4 = a \cdot 2 \cdot (-2) = -4a\), so \(a = 1\). 3. Expand: \(f(x) = (x + 1)(x - 3) = x^2 - 2x - 3\).

\(f(x) = x^2 - 2x - 3\)

- Read each maximum directly from vertex form. - Set each function equal to \(0\) to find where the jump begins and ends. - How does the absolute value of the leading coefficient affect the width of a parabola?

1. The function \(f\) has vertex \((2, 2)\), so its maximum height is \(2\,\text{m}\). The function \(g\) has vertex \((3, 1.8)\), so its maximum height is \(1.8\,\text{m}\). Therefore, the first jump reaches the greater maximum height. 2. For \(f\), solve \(0 = -0.5(x - 2)^2 + 2\). Then \((x - 2)^2 = 4\), so \(x = 0\) or \(x = 4\). The first jump covers \(4\,\text{m}\). 3. For \(g\), solve \(0 = -0.2(x - 3)^2 + 1.8\). Then \((x - 3)^2 = 9\), so \(x = 0\) or \(x = 6\). The second jump covers \(6\,\text{m}\). 4. The second jump is flatter because its leading coefficient has the smaller absolute value: \(|-0.2| < |-0.5|\). It also has a lower maximum height while covering a greater horizontal distance.

a) The first jump, with a maximum height of \(2\,\text{m}\), compared with \(1.8\,\text{m}\) for the second jump b) First jump: \(4\,\text{m}\) Second jump: \(6\,\text{m}\) c) The second jump is flatter because \(|-0.2| < |-0.5|\); it is wider and has a lower maximum.

- Which form of a quadratic is useful when its zeros are known? - Where is the axis of symmetry relative to the two zeros? - How can the given point help you determine the leading coefficient?

1. Use intercept form: \(f(x) = a(x + 2)(x - 6)\). 2. Substitute \(P(0, 3)\): \(3 = a(0 + 2)(0 - 6) = -12a\), so \(a = -0.25\). 3. The axis of symmetry is halfway between the zeros: \(x = \frac{-2 + 6}{2} = 2\). 4. Evaluate the function at \(x = 2\): \(f(2) = -0.25(2 + 2)(2 - 6) = -0.25 \cdot 4 \cdot (-4) = 4\). 5. Therefore, the vertex is \((2, 4)\).

The vertex is \((2, 4)\).

- Use the highest point as the vertex in vertex form. - Substitute the release point to determine the leading coefficient. - Hitting the ground means the height is \(0\).

1. Use the vertex \((8, 12)\): \(f(x) = a(x - 8)^2 + 12\). 2. Substitute the release point \((0, 6)\): \(6 = a(0 - 8)^2 + 12\). Then \(64a = -6\), so \(a = -\frac{3}{32}\). Thus, \(f(x) = -\frac{3}{32}(x - 8)^2 + 12\). 3. Evaluate at \(x = 12\): \(f(12) = -\frac{3}{32} \cdot (12 - 8)^2 + 12 = 10.5\). The ball is \(10.5\,\text{ft}\) high. 4. Set \(f(x) = 0\): \(0 = -\frac{3}{32}(x - 8)^2 + 12\). Then \((x - 8)^2 = 128\), so \(x = 8 \pm 8\sqrt{2}\). 5. The context-valid positive distance is \(x = 8 + 8\sqrt{2} \approx 19.31\,\text{ft}\).

a) \(f(x) = -\frac{3}{32}(x - 8)^2 + 12\) b) \(10.5\,\text{ft}\) c) \(8 + 8\sqrt{2}\,\text{ft} \approx 19.31\,\text{ft}\)

- Complete the square to rewrite the quadratic in vertex form. - Substitute \(-3 + h\) and \(-3 - h\) for \(x\). - Expand each squared binomial carefully. - Compare the simplified results.

1. Complete the square: \(f(x) = x^2 + 6x + 5 = (x + 3)^2 - 4\). Therefore, the axis of symmetry is \(x = -3\). 2. Evaluate the function on one side of the axis: \(f(-3 + h) = (-3 + h)^2 + 6(-3 + h) + 5 = h^2 - 4\). 3. Evaluate the function the same distance on the other side: \(f(-3 - h) = (-3 - h)^2 + 6(-3 - h) + 5 = h^2 - 4\). 4. Since \(f(-3 + h) = f(-3 - h)\) for every real \(h\), the graph is symmetric about \(x = -3\).

Vertex form: \(f(x) = (x + 3)^2 - 4\) Axis of symmetry: \(x = -3\) The verification gives \(f(-3 + h) = f(-3 - h) = h^2 - 4\).

- Use each zero to write a corresponding linear factor. - Substitute the coordinates of the given point to determine \(a\). - Multiply the binomials and distribute \(a\) to obtain standard form.

1. Zeros of \(-2\) and \(6\) give the factors \(x + 2\) and \(x - 6\), so \(f(x) = a(x + 2)(x - 6)\). Thus, \(b = -2\) and \(c = 6\), or the two values may be reversed. 2. Use the point \((2, 8)\): \(8 = a(2 + 2)(2 - 6) = a(4)(-4) = -16a\). 3. Solve for the leading coefficient: \(a = -\frac{1}{2}\). 4. The intercept form is \(f(x) = -\frac{1}{2}(x + 2)(x - 6)\). 5. Expand: \((x + 2)(x - 6) = x^2 - 4x - 12\), so \(f(x) = -\frac{1}{2}x^2 + 2x + 6\).

\(a = -\frac{1}{2}\), with \((b, c) = (-2, 6)\) or \((6, -2)\) Standard form: \(f(x) = -\frac{1}{2}x^2 + 2x + 6\)

- When a parabola touches the x-axis at a zero, that zero has multiplicity 2. - A double zero means the two factors \(x - s\) and \(x - t\) are the same. - The y-intercept occurs where \(x = 0\).

1. Because the graph touches the x-axis at \(x = 3\), \(x = 3\) is a double zero. Therefore, \(s = t = 3\), and \(g(x) = a(x - 3)^2\). 2. Substitute the point \((5, -10)\): \(-10 = a(5 - 3)^2 = 4a\). Thus, \(a = -\frac{5}{2}\). 3. To find the y-intercept, evaluate \(g(0)\): \(g(0) = -\frac{5}{2}(0 - 3)^2 = -\frac{5}{2}(9) = -\frac{45}{2}\). 4. The y-intercept is \((0, -22.5)\).

1. \(a = -\frac{5}{2}\), \(s = 3\), and \(t = 3\) 2. \((0, -22.5)\)

- Can you factor a negative sign from the entire expression? - What does the leading coefficient tell you about the opening direction? - What is the \(x\)-coordinate of a point on the \(y\)-axis? - What do you know about the value of a square?

1. Factor out \(-1\): \(h(x) = -(x^2 - 10x + 25) = -(x - 5)^2\). The vertex is \((5, 0)\), and the negative leading coefficient means the parabola opens downward. 2. Evaluate at \(x = 0\): \(h(0) = -25\), so the \(y\)-intercept is \((0, -25)\). 3. For every real \(x\), \((x - 5)^2 \geq 0\). Therefore, \(-(x - 5)^2 \leq 0\), so \(h(x)\) can never be positive.

a) Vertex \((5, 0)\); the parabola opens downward. b) \((0, -25)\) c) Since \(h(x) = -(x - 5)^2\) and every square is nonnegative, \(h(x) \leq 0\) for all real \(x\).

- Complete the square to rewrite the function in vertex form. - Compare the vertex's y-coordinate with \(0\). - The solutions of \(f(x) = 0\) are the x-coordinates of the x-intercepts.

1. Complete the square: \(f(x) = x^2 - 6x + 5 = (x - 3)^2 - 4\). The vertex is \((3, -4)\). 2. The leading coefficient is positive, so the parabola opens upward. Its minimum value is \(-4\), which is below the x-axis, so the graph crosses the x-axis twice. 3. Set \((x - 3)^2 - 4 = 0\): \((x - 3)^2 = 4\), so \(x = 1\) or \(x = 5\). The x-intercepts are \((1, 0)\) and \((5, 0)\).

1. Vertex form: \(f(x) = (x - 3)^2 - 4\); vertex: \((3, -4)\) 2. The parabola opens upward and its vertex is below the x-axis, so it has two x-intercepts. 3. The x-intercepts are \((1, 0)\) and \((5, 0)\).

- Find the vertex of the parabola. - A horizontal tangent passes through the vertex. - Compare \(c = 5\) with the maximum function value.

1. Write the parabola in vertex form: \(p(x) = -(x^2 - 4x) = -(x - 2)^2 + 4\). 2. The vertex is \((2, 4)\), and the parabola opens downward. 3. A horizontal tangent occurs at the vertex, so \(c = 4\), and the point of tangency is \((2, 4)\). 4. The maximum value of the parabola is \(4\). Therefore, the line \(y = 5\) lies above the entire parabola and has no intersection points.

a) \(c = 4\) b) \((2, 4)\) c) No intersections

- Use the x-coordinate \(k\) in the given vertex-locus function. - Recall the vertex form \(y=(x-h)^2+v\). - A parabola with one x-intercept is tangent to the x-axis at its vertex.

1. If the x-coordinate of the vertex is \(k\), then its y-coordinate is \(g(k)=k^2-6\). Thus, the vertex is \((k, k^2-6)\). 2. A parabola with leading coefficient \(1\) and vertex \((h, v)\) has equation \(y=(x-h)^2+v\). Therefore, \(f_k(x)=(x-k)^2+k^2-6\). 3. The parabola has exactly one x-intercept when its vertex lies on the x-axis. Hence, \(k^2-6=0\), so \(k=\pm\sqrt6\).

a) \(f_k(x)=(x-k)^2+k^2-6\) b) \(k=\sqrt6\) or \(k=-\sqrt6\)

- How can you find the vertex of a quadratic in standard form? - What value of \(x\) gives the \(y\)-intercept? - What equation gives the \(x\)-intercepts? - What does the leading coefficient tell you about the graph?

1. The vertex has \(x\)-coordinate \(x = -\frac{-8}{2 \cdot 2} = 2\). Then \(f(2) = 2 \cdot 2^2 - 8 \cdot 2 + 6 = -2\), so the vertex is \((2, -2)\). 2. The \(y\)-intercept is \((0, f(0)) = (0, 6)\). 3. For the \(x\)-intercepts, solve \(2x^2 - 8x + 6 = 0\). Divide by \(2\): \(x^2 - 4x + 3 = 0\). Factoring gives \((x - 1)(x - 3) = 0\), so the intercepts are \((1, 0)\) and \((3, 0)\). 4. Since the leading coefficient is positive, the parabola opens upward. Since \(|2| > 1\), it is narrower than \(y = x^2\).

a) \((2, -2)\) b) \(x\)-intercepts: \((1, 0)\) and \((3, 0)\); \(y\)-intercept: \((0, 6)\) c) It opens upward and is narrower than \(y = x^2\).

- Which form lets you read a vertex directly? - Use the vertex's position and the opening direction to reason about \(x\)-intercepts. - How many intercepts occur when the vertex lies exactly on the \(x\)-axis? - What input gives the \(y\)-intercept?

1. From vertex form, \(p_1\) has vertex \((-2, -1)\). 2. Rewrite \(p_2\): \(p_2(x) = -0.5(x^2 - 4x + 4) = -0.5(x - 2)^2\). Its vertex is \((2, 0)\). 3. The graph of \(p_1\) opens upward and has its vertex below the \(x\)-axis, so it crosses the \(x\)-axis twice. The graph of \(p_2\) has its vertex on the \(x\)-axis, so it touches the axis once. Therefore, \(p_1\) has more \(x\)-intercepts. 4. Evaluate \(p_1(0) = (0 + 2)^2 - 1 = 3\), so the \(y\)-intercept is \((0, 3)\).

a) \(p_1\): \((-2, -1)\); \(p_2\): \((2, 0)\) b) \(p_1\) has two \(x\)-intercepts, while \(p_2\) has one, so \(p_1\) has more. c) \((0, 3)\)

- Can you rewrite each function in vertex form? - What equation must an intersection point satisfy for both functions? - After finding each \(x\)-coordinate, how do you find the corresponding \(y\)-coordinate? - What does the leading coefficient tell you about opening and width?

1. Rewrite the functions: \(p_1(x) = (x - 2)^2 - 1\), so its vertex is \((2, -1)\). Also, \(p_2(x) = -(x - 2)^2 + 3\), so its vertex is \((2, 3)\). 2. Set the equations equal: \(x^2 - 4x + 3 = -x^2 + 4x - 1\). This simplifies to \(x^2 - 4x + 2 = 0\), so \(x = 2 \pm \sqrt{2}\). 3. Substituting either value into \(p_1(x) = (x - 2)^2 - 1\) gives \(y = 1\). The intersections are \((2 - \sqrt{2}, 1)\) and \((2 + \sqrt{2}, 1)\). 4. Both leading coefficients have absolute value \(1\), so the parabolas have the same width as \(y = x^2\). The first opens upward, and the second opens downward.

1) \((2, -1)\) and \((2, 3)\) 2) \((2 - \sqrt{2}, 1)\) and \((2 + \sqrt{2}, 1)\) 3) They have the same width; \(p_1\) opens upward and \(p_2\) opens downward.

- Which form of a quadratic function uses its zeros directly? - Substitute the coordinates of the given point to determine the leading coefficient. - Use the distributive property to expand the product of two binomials.

1. Use the zeros to write \(g(x) = a(x + 1)(x - 3)\). 2. Substitute the point \(P(2, -6)\): \(-6 = a(2 + 1)(2 - 3) = a \cdot 3 \cdot (-1) = -3a\). Therefore, \(a = 2\). 3. The factored form is \(g(x) = 2(x + 1)(x - 3)\). 4. Expand the factors: \((x + 1)(x - 3) = x^2 - 2x - 3\). 5. Multiply by \(2\): \(g(x) = 2x^2 - 4x - 6\).

a) \(g(x) = 2(x + 1)(x - 3)\) b) \(g(x) = 2x^2 - 4x - 6\)

- Consider what happens if the coefficient of \(x^2\) equals zero. - The sign of \(a\) determines whether the graph opens upward or downward. - Clear the fraction before solving the quadratic equation.

1. A quadratic function requires \(a\ne0\); otherwise, the squared term disappears. If \(a>0\), the parabola opens upward, and if \(a<0\), it opens downward. Larger values of \(|a|\) make the parabola narrower, while values \(0<|a|<1\) make it wider. 2. Set the function equal to zero: \(\frac{1}{2}x^2-2x-6=0\). Multiply by \(2\) to get \(x^2-4x-12=0\). 3. Factor: \((x-6)(x+2)=0\). Therefore, the zeros are \(x=-2\) and \(x=6\).

\(a\ne0\). Its sign determines the opening direction, and \(|a|\) affects the width. The zeros are \(x=-2\) and \(x=6\).

- In \(a(x-h)^2+k\), the vertex is \((h, k)\). - Expand the square and combine like terms. - After setting the function equal to zero, look for a common factor.

1. The equation is in vertex form \(a(x-h)^2+k\), so the vertex is \((3, 18)\). 2. Expand: \(g(x)=-2(x^2-6x+9)+18=-2x^2+12x\). 3. Set \(g(x)=0\): \(-2x^2+12x=-2x(x-6)=0\). The zeros are \(x=0\) and \(x=6\).

1. Vertex form; vertex \((3, 18)\) 2. \(g(x)=-2x^2+12x\) 3. \(x=0\) and \(x=6\)

- Simplify the quadratic equation before solving. - The axis of symmetry is halfway between the zeros. - Use vertex form to construct a parabola with the same vertex. - To avoid real zeros, keep the entire graph on one side of the x-axis.

1. Set \(f(x)=0\) and divide by \(2\): \(x^2+6x+7=0\). 2. Complete the square or use the quadratic formula: \(x=-3\pm\sqrt{2}\). These are the zeros. 3. The axis of symmetry is midway between the zeros, so \(x=-3\). Evaluate \(f(-3)=18-36+14=-4\). The vertex is \((-3, -4)\). 4. One example is \(g(x)=-(x+3)^2-4\). It has the same vertex and opens downward. Its maximum value is \(-4\), so its graph stays below the x-axis and has no real zeros.

a) \(x=-3-\sqrt{2}\) and \(x=-3+\sqrt{2}\) b) Vertex: \((-3, -4)\). One example is \(g(x)=-(x+3)^2-4\).

- Complete the square to identify the vertex. - Replace the parameter with the vertex x-coordinate. - Solve the inequality for the vertex y-coordinate.

1. Complete the square: \(f_a(x)=(x-a)^2-a^2+3a\). Therefore, the vertex is \((a, -a^2+3a)\). 2. Since the vertex x-coordinate is \(x=a\), the locus is \(y=-x^2+3x\). 3. The vertex is above the x-axis when \(-a^2+3a>0\), or \(a(3-a)>0\). This holds for \(0<a<3\).

a) \((a, -a^2+3a)\) b) \(y=-x^2+3x\) c) \(0<a<3\)

- Complete the square to identify the vertex. - Replace \(k\) with the vertex x-coordinate for the locus. - Factor the quadratic inequality.

1. Complete the square: \(g_k(x)=-(x-k)^2+k^2-4k\). Therefore, the vertex is \((k, k^2-4k)\). 2. Since the vertex x-coordinate is \(x=k\), its y-coordinate is \(y=x^2-4x\), proving the locus. 3. The vertex lies below \(y=-3\) when \(k^2-4k<-3\), or \((k-1)(k-3)<0\). Thus \(1<k<3\).

a) \((k, k^2-4k)\) b) \(y=x^2-4x\) c) \(1<k<3\)

- Identify the highest point of each curve. - Substitute a clearly readable ground-level point into vertex form to find the leading coefficient. - At an intersection, the two functions have equal outputs. - Keep only solutions that lie in the physical domain shown.

1. From the graph, the vertices are \((2, 4)\) for \(f\) and \((3, 3)\) for \(g\). 2. Write \(f(x) = a(x - 2)^2 + 4\). The graph shows that \((0, 0)\) lies on \(f\), so \(0 = 4a + 4\), giving \(a = -1\). Thus, \(f(x) = -(x - 2)^2 + 4\). 3. Write \(g(x) = a(x - 3)^2 + 3\). The graph shows that \((1, 0)\) lies on \(g\), so \(0 = 4a + 3\), giving \(a = -0.75\). Thus, \(g(x) = -0.75(x - 3)^2 + 3\). 4. Set the functions equal: \(-(x - 2)^2 + 4 = -0.75(x - 3)^2 + 3\). Simplifying gives \(x^2 + 2x - 15 = 0\), or \((x - 3)(x + 5) = 0\). 5. The solutions are \(x = 3\) and \(x = -5\). Only \(x = 3\) lies in the physical domain shown. Since \(f(3) = 3\), the intersection point is \((3, 3)\).

a) \(f\): \((2, 4)\) \(g\): \((3, 3)\) b) \(f(x) = -(x - 2)^2 + 4\) \(g(x) = -0.75(x - 3)^2 + 3\) c) \((3, 3)\)

- Read the highest point and the y-intercept from the graph. - Substitute the release point into vertex form to find the leading coefficient. - Test the hoop by evaluating the function at its horizontal position. - Reaching the ground means the height is \(0\).

1. The graph shows vertex \((12, 10.5)\) and release point \((0, 6)\). Begin with \(f(x) = a(x - 12)^2 + 10.5\). 2. Substitute \((0, 6)\): \(6 = a(0 - 12)^2 + 10.5\). Then \(144a = -4.5\), so \(a = -0.03125 = -\frac{1}{32}\). Therefore, \(f(x) = -\frac{1}{32}(x - 12)^2 + 10.5\). 3. Evaluate the height at the hoop: \(f(16) = -\frac{1}{32} \cdot (16 - 12)^2 + 10.5 = 10\). The ball passes exactly through the hoop. 4. Set the height equal to \(0\): \(0 = -\frac{1}{32}(x - 12)^2 + 10.5\). Then \((x - 12)^2 = 336\), so \(x = 12 \pm 4\sqrt{21}\). 5. The context-valid solution is \(x = 12 + 4\sqrt{21} \approx 30.33\,\text{ft}\).

a) Vertex: \((12, 10.5)\); release height: \(6\,\text{ft}\) \(f(x) = -\frac{1}{32}(x - 12)^2 + 10.5\) b) Yes. \(f(16) = 10\,\text{ft}\). c) \(12 + 4\sqrt{21}\,\text{ft} \approx 30.33\,\text{ft}\)

- Isolate the factor multiplied by \(k\). - Substitute the coordinates of \(C\) into the function. - Use the formula for the x-coordinate of a parabola’s vertex. - Match the resulting function to a labeled graph.

1. Rewrite the function as \(f_k(x)=k(x^2-4x+3)-x+3\). The value is independent of \(k\) when \(x^2-4x+3=(x-1)(x-3)=0\). Therefore, the common points are \(A(1, 2)\) and \(B(3, 0)\). 2. Evaluate at \(x=2\): \(f_k(2)=1-k\). Setting \(1-k=-2\) gives \(k=3\). 3. The vertex x-coordinate is \(x_v=\frac{4k+1}{2k}=2+\frac{1}{2k}\). Setting this equal to \(2.5\) gives \(k=1\). The graph for \(k=1\) is \(y=x^2-5x+6\), which is labeled q.

a) \(A(1, 2)\) and \(B(3, 0)\) b) \(k=3\) c) \(k=1\); graph q

- Use the two x-intercepts to write a factored quadratic. - Use one labeled graph to determine the leading coefficient. - The vertex lies midway between the zeros. - Eliminate \(k\) and keep its restriction.

1. Each parabola opens downward and has zeros at \(x=0\) and \(x=k\). Thus it has the form \(f_k(x)=c x(x-k)\). 2. The graph for \(k=2\) has vertex \((1, 1)\). Substitution gives \(1=c(1)(1-2)\), so \(c=-1\). Therefore, \(f_k(x)=-x^2+kx\). 3. The vertex lies midway between the zeros, so \(x=\frac{k}{2}\). Its y-coordinate is \(f_k\left(\frac{k}{2}\right)=\frac{k^2}{4}\). Thus the vertex is \(\left(\frac{k}{2}, \frac{k^2}{4}\right)\). 4. Since \(k=2x\), the vertex y-coordinate becomes \(y=x^2\). Because \(k\neq0\), the attainable x-values satisfy \(x\neq0\).

a) \(f_k(x)=-x^2+kx\) b) \(\left(\frac{k}{2}, \frac{k^2}{4}\right)\) c) \(y=x^2\) for \(x\neq0\)

- Read both x-intercepts from the graph. - Write the factors \((x - x_1)(x - x_2)\). - Expand and combine like terms.

1. The graph crosses the x-axis at \(x_1 = -2\) and \(x_2 = 1\). 2. Because the quadratic is monic, its factored form is \(f(x) = (x - x_1)(x - x_2) = (x + 2)(x - 1)\). 3. Expanding gives \(f(x) = x^2 - x + 2x - 2 = x^2 + x - 2\).

\(f(x) = x^2 + x - 2\)

- Estimate where the graph crosses the x-axis. - Set the function equal to \(0\) for exact values. - Use both square roots, then evaluate the radicals with a calculator.

1. From the graph, the zeros are approximately \(x \approx -2.7\) and \(x \approx 0.7\). 2. Set the function equal to zero: \((x + 1)^2 - 3 = 0\). 3. Then \((x + 1)^2 = 3\), so \(x + 1 = \pm\sqrt{3}\). 4. Therefore, \(x = -1 + \sqrt{3}\) or \(x = -1 - \sqrt{3}\). 5. Rounded to the nearest hundredth, the values are \(0.73\) and \(-2.73\).

a) Approximately \(x = -2.7\) and \(x = 0.7\) b) \(x = -1 - \sqrt{3} \approx -2.73\) and \(x = -1 + \sqrt{3} \approx 0.73\)

- Interpret the x-intercepts as the points where the tunnel meets the roadway. - The constant term in \(y = ax^2 + c\) is the height at \(x = 0\). - A centered vehicle extends half its width to each side of the y-axis.

1. From the graph, the zeros are \(x = -2\) and \(x = 2\), and the vertex is \((0, 4)\). The ground-level width is \(2 - (-2) = 4\,\text{m}\). 2. In \(y = ax^2 + c\), the vertex \((0, 4)\) gives \(c = 4\). Using the zero \((2, 0)\), \(0 = a \cdot 2^2 + 4\), so \(4a = -4\) and \(a = -1\). Thus, \(h(x) = -x^2 + 4\). 3. A centered \(2\,\text{m}\)-wide vehicle extends from \(x = -1\) to \(x = 1\). The critical clearance is at its outer top corners: \(h(1) = -1^2 + 4 = 3\). The vehicle can be at most \(3\,\text{m}\) high.

a) Zeros: \(x = -2\) and \(x = 2\); vertex: \((0, 4)\); width: \(4\,\text{m}\) b) \(h(x) = -x^2 + 4\) c) \(3\,\text{m}\)

- Use symmetry to identify the vertex. - Determine the opening direction from the three points. - Decide where the vertex must be for two x-intercepts.

1. Points \(A\) and \(C\) are symmetric about the y-axis and have the same y-coordinate, so \(B(0, 1)\) is the vertex. 2. The parabola opens upward and its vertex lies above the x-axis. Therefore, it has no real zeros and \(D < 0\). 3. To make \(D > 0\), the vertex must lie below the x-axis. The y-coordinate of \(B\) must be decreased from \(1\) to a value less than \(0\). 4. Therefore, it must be decreased by more than \(1\).

a) \(D < 0\) b) The y-coordinate must be decreased by more than \(1\).

- Read the vertex from the graph. - Use another labeled point to determine \(a\). - Set the vertex-form expression equal to \(0\) to find the zeros.

1. The graph shows the vertex \((-2, -4.5)\) and the point \((1, 0)\). 2. Write \(f(x) = a(x + 2)^2 - 4.5\). 3. Substitute \((1, 0)\): \(0 = a(1 + 2)^2 - 4.5\), so \(9a = 4.5\) and \(a = 0.5\). 4. Therefore, \(f(x) = 0.5(x + 2)^2 - 4.5\). 5. Set \(f(x) = 0\): \(0.5(x + 2)^2 = 4.5\), so \((x + 2)^2 = 9\). 6. Thus, \(x + 2 = \pm 3\), giving \(x = 1\) or \(x = -5\).

a) \(f(x) = 0.5(x + 2)^2 - 4.5\) b) \(x = -5\) and \(x = 1\)

- Read each vertex and one additional point from the graph. - Use vertex form to determine the leading coefficient. - Set the two function rules equal and factor the resulting equation.

1. The downward-opening parabola \(f\) has vertex \((0, 5)\) and passes through \((1, 4)\). Thus, \(f(x) = ax^2 + 5\), and \(4 = a + 5\), so \(a = -1\). Therefore, \(f(x) = -x^2 + 5\). 2. The upward-opening parabola \(g\) has vertex \((2, 1)\) and passes through \((1, 2)\). Thus, \(g(x) = a(x - 2)^2 + 1\), and \(2 = a + 1\), so \(a = 1\). Therefore, \(g(x) = (x - 2)^2 + 1\). 3. Set the functions equal: \(-x^2 + 5 = (x - 2)^2 + 1\). 4. Simplify: \(2x^2 - 4x = 0\), so \(2x(x - 2) = 0\). Thus, \(x = 0\) or \(x = 2\). 5. The corresponding y-values are \(5\) and \(1\), giving the intersection points \((0, 5)\) and \((2, 1)\).

a) \(f(x) = -x^2 + 5\) and \(g(x) = (x - 2)^2 + 1\) b) \((0, 5)\) and \((2, 1)\)

- Use the vertex coordinates in the given locus equation. - Substitute the given point into the family equation. - Expand carefully using the binomial formula.

1. The vertex is \((a, c_a)\). Since it lies on \(y=0.25x^2\), \(c_a=0.25a^2\). 2. Therefore, \(f_a(x)=-0.5(x-a)^2+0.25a^2\). 3. Substitute \(P(2, 2)\): \(2=-0.5(2-a)^2+0.25a^2\). 4. Simplifying gives \(a^2-8a+16=0\), or \((a-4)^2=0\). Thus \(a=4\).

a) \(f_a(x)=-0.5(x-a)^2+0.25a^2\) b) \(a=4\)

- Isolate the term containing \(k\). - Complete the square to identify the vertex. - Replace \(k\) with the vertex x-coordinate. - Use the discriminant to determine the number of real zeros.

1. Rewrite the function as \(f_k(x)=x^2+3-2k(x-1)\). At \(x=1\), the parameter term is zero, so \(f_k(1)=4\). 2. Complete the square: \(f_k(x)=(x-k)^2-k^2+2k+3\). Therefore, the vertex is \((k, -k^2+2k+3)\). 3. Since the vertex x-coordinate is \(x=k\), the locus is \(y=-x^2+2x+3\). 4. The discriminant of \(x^2-2kx+2k+3=0\) is \(4(k-3)(k+1)\). There are no real zeros when the discriminant is negative, so \(-1<k<3\). 5. The parabola is tangent to the x-axis when the discriminant is zero, giving \(k=-1\) or \(k=3\).

a) \(P(1, 4)\) b) \((k, -k^2+2k+3)\) c) \(y=-x^2+2x+3\) d) \(-1<k<3\) e) \(k=-1\) or \(k=3\)

- Write the rectangle’s area as the product of its width and height. - Use the line equation to express the height in terms of the width. - Maximize the resulting downward-opening quadratic by finding its vertex.

1. The rectangle has area \(A = uv\). 2. Since \(P\) lies on \(g\), \(v = -0.75u + 6\). 3. Substitute to obtain \(A(u) = u(-0.75u + 6) = -0.75u^2 + 6u\). 4. This parabola opens downward. Its vertex occurs at \(u = -\frac{6}{2(-0.75)} = 4\). 5. Then \(v = -0.75(4) + 6 = 3\). 6. The maximum area is \(A(4) = 4 \cdot 3 = 12\,\text{ft}^2\).

\(P = (4, 3)\); maximum area \(12\,\text{ft}^2\)

- First read the common y-intercept from the graph. - Identify the coordinates of the second intersection. - Substitute a known point into the function equation to determine the missing coefficient.

1. The shared y-intercept shown in the graph is \(6\), so \(c=6\). 2. The second intersection is the x-intercept \((-3,0)\). Because this point lies on \(f\), \(f(-3)=0\). 3. Substitute \(x=-3\) and \(c=6\): \(0=(-3)^2-3b+6\). 4. Simplify: \(0=15-3b\), so \(3b=15\) and \(b=5\).

\(b=5\)

- Use the intersection on the y-axis to determine \(k\). - Read the coordinates of the second intersection from the graph. - Substitute that point into the parabola equation to determine \(a\).

1. The graphs intersect on the y-axis at \((0,3)\). Since \(g(0)=k\), it follows that \(k=3\), so \(g(x)=-x+3\). 2. The graph shows a second intersection at \((1,2)\). 3. Because this point lies on the parabola, substitute \(x=1\) and \(y=2\): \(2=a-4+3\). 4. Simplify: \(2=a-1\), so \(a=3\).

\(a=3\)

- Where is the axis of symmetry relative to the two zeros? - How can the vertex help determine the unknown leading coefficient in intercept form? - After finding the leading coefficient, expand the factors carefully.

1. The axis of symmetry is halfway between the zeros: \(x = \frac{-3 + 1}{2} = -1\). Therefore, the vertex is \((-1, 4)\). 2. Write the function in intercept form: \(f(x) = a(x + 3)(x - 1)\). 3. Substitute the vertex: \(4 = a \cdot 2 \cdot (-2) = -4a\), so \(a = -1\). 4. Expand: \(f(x) = -(x + 3)(x - 1) = -x^2 - 2x + 3\).

\(f(x) = -x^2 - 2x + 3\)

- For two points with the same y-coordinate, where is the axis of symmetry relative to their x-coordinates? - After finding the axis of symmetry, substitute one point into vertex form. - Check whether the vertex lies in the restricted domain. - Compare the function values at the vertex and at both endpoints.

1. The points \(P\) and \(Q\) have the same y-coordinate, so their x-coordinates are equally far from the axis of symmetry. The axis is halfway between them: \(h=\frac{-2+4}{2}=1\). 2. Use \(f(x)=(x-1)^2+k\) and substitute \(Q(4, 6)\): \(6=(4-1)^2+k=9+k\), so \(k=-3\). The vertex is \((1, -3)\). 3. Expand the vertex form: \(f(x)=(x-1)^2-3=x^2-2x-2\). 4. The vertex x-coordinate \(1\) lies in \([-3, 3]\), so the minimum value is \(-3\). 5. Evaluate the endpoints: \(f(-3)=(-3-1)^2-3=13\) and \(f(3)=(3-1)^2-3=1\). 6. The maximum value is \(13\), so the range on the restricted domain is \([-3, 13]\).

a) The vertex is \((1, -3)\). b) \(f(x)=x^2-2x-2\) c) \([-3, 13]\)

- A point on a graph must make the function equation true. - Taking square roots creates two possible cases. - Look for a value of \(h\) that works for several points.

1. A point \((x, y)\) on the graph satisfies \(y=(x-h)^2\), so \(x-h=\sqrt{y}\) or \(x-h=-\sqrt{y}\). 2. For \(A(-3, 4)\), \(-3-h=2\) or \(-3-h=-2\), giving \(h=-5\) or \(h=-1\). 3. For \(B(0, 1)\), \(-h=1\) or \(-h=-1\), giving \(h=-1\) or \(h=1\). The common value for \(A\) and \(B\) is \(h=-1\). 4. Check \(C(-2, 1)\): \(1=(-2-(-1))^2=1\). Thus, \(A\), \(B\), and \(C\) lie on \(f(x)=(x+1)^2\). 5. Point \(F(2, 0)\) gives \(0=(2-h)^2\), so \(h=2\). 6. Check the remaining points: \((1-2)^2=1\) for \(D\), and \((4-2)^2=4\) for \(E\). Thus, \(D\), \(E\), and \(F\) lie on \(f(x)=(x-2)^2\).

The two values are \(h=-1\) and \(h=2\). For \(h=-1\), the function is \(f(x)=(x+1)^2\), and the points are \(A\), \(B\), and \(C\). For \(h=2\), the function is \(f(x)=(x-2)^2\), and the points are \(D\), \(E\), and \(F\).

- If the vertex lies on the x-axis, what is the vertex's y-coordinate? - Compare the situation with vertex form \(f(x)=(x-h)^2+k\). - How many real solutions can an equation of the form \(u^2=a\) have when \(a>0\)? - Can the square of a real number be negative?

1. A translation of \(y=x^2\) with its vertex on the x-axis has the form \(f(x)=(x-h)^2\). 2. For part a, substitute \(A(1, 4)\): \(4=(1-h)^2\). 3. Taking both square roots gives \(1-h=2\) or \(1-h=-2\). Thus, \(h=-1\) or \(h=3\). 4. The possible equations are \(f_1(x)=(x+1)^2\) and \(f_2(x)=(x-3)^2\). 5. For part b, substituting \(B(1, -4)\) gives \(-4=(1-h)^2\). 6. The square of a real number cannot be negative, so there is no real value of \(h\) and no parabola of the required form through \(B\).

a) \(f_1(x)=(x+1)^2\) and \(f_2(x)=(x-3)^2\) b) Every function of the form \(f(x)=(x-h)^2\) has \(f(x)\ge 0\) for all real \(x\), so it cannot contain a point with y-coordinate \(-4\).

- What does the line \(y=2\) tell you about the vertex's y-coordinate? - Substitute the given point into vertex form. - What function value is used to find an x-intercept? - Consider the minimum value of each upward-opening parabola.

1. Since the vertex lies on \(y=2\), the vertex form is \(f(x)=(x-h)^2+2\). 2. Substitute \(P(4, 6)\): \(6=(4-h)^2+2\), so \(4=(4-h)^2\). 3. Taking both square roots gives \(4-h=2\) or \(4-h=-2\). Thus, \(h=2\) or \(h=6\). 4. The two equations are \(f_1(x)=(x-2)^2+2\) and \(f_2(x)=(x-6)^2+2\). 5. For an x-intercept, set \(f(x)=0\): \((x-h)^2+2=0\), so \((x-h)^2=-2\). 6. A square cannot be negative for real \(x\), so neither parabola intersects the x-axis. Equivalently, both parabolas open upward and have minimum value \(2\).

1. \(f_1(x)=(x-2)^2+2\) and \(f_2(x)=(x-6)^2+2\) 2. Neither parabola intersects the x-axis. Setting either function equal to \(0\) gives a squared expression equal to \(-2\), which has no real solution.

- What is the vertex's y-coordinate when the vertex lies on the x-axis? - Write the general vertex form for this situation. - Substitute the y-intercept into the equation. - Remember that taking square roots produces two cases for a positive value.

1. Since the vertex lies on the x-axis, the vertex form is \(f(x)=(x-h)^2\). 2. The y-intercept gives \(f(0)=6.25\), so \((0-h)^2=6.25\) and \(h^2=6.25\). 3. Taking both square roots gives \(h=2.5\) or \(h=-2.5\). 4. For \(h=2.5\), \(f_1(x)=(x-2.5)^2=x^2-5x+6.25\). 5. For \(h=-2.5\), \(f_2(x)=(x+2.5)^2=x^2+5x+6.25\). 6. There are two solutions because \(2.5^2=(-2.5)^2=6.25\), so the vertex may lie the same distance to either side of the y-axis.

The two equations are \(f_1(x)=x^2-5x+6.25\) and \(f_2(x)=x^2+5x+6.25\). There are two solutions because the vertex can be at \(x=2.5\) or \(x=-2.5\), and both positions give the same y-intercept.

- Where is the axis of symmetry when two points have the same y-coordinate? - Write the general vertex form of a translation of \(y=x^2\). - Once the axis is known, how many unknown parameters remain?

1. Write \(g(x)=(x-h)^2+k\). 2. The points \(A\) and \(B\) have the same y-coordinate, so they are symmetric across the axis of the parabola. The axis lies halfway between their x-coordinates: \(h=\frac{-2+4}{2}=1\). 3. Substitute \(B(4, 3)\): \(3=(4-1)^2+k=9+k\), so \(k=-6\). 4. Therefore, \(g(x)=(x-1)^2-6\). 5. The equal y-coordinates determine the axis \(x=1\) uniquely. Since the leading coefficient is fixed at \(1\), substituting either point then determines \(k\) uniquely. Thus, there is exactly one such function.

a) \(g(x)=(x-1)^2-6\) b) The equal y-coordinates force the axis of symmetry to be \(x=1\). With leading coefficient \(1\), either point then determines the vertical shift uniquely, so only one function satisfies the conditions.

- Evaluate the function values in the two conditions first. - Use the general vertex form of a translated parent parabola. - How can subtracting the two equations eliminate the vertical-shift parameter? - To test a point, compare its y-coordinate with the function value at its x-coordinate.

1. Evaluate the given function values: \(f(3)=3^2=9\) and \(f(1)=1^2=1\). 2. Write \(g(x)=(x-h)^2+k\). The conditions give \((1-h)^2+k=9\) and \((3-h)^2+k=1\). 3. Subtract the second equation from the first: \((1-h)^2-(3-h)^2=8\). 4. Expand and simplify: \((1-2h+h^2)-(9-6h+h^2)=8\), so \(4h-8=8\) and \(h=4\). 5. Substitute \(h=4\) into \((3-h)^2+k=1\): \((3-4)^2+k=1\), so \(k=0\). 6. Therefore, \(g(x)=(x-4)^2\). 7. Check \(P\): \(g(2)=(2-4)^2=4\). Since \(4\ne 0\), \(P(2, 0)\) is not on the graph.

The equation is \(g(x)=(x-4)^2\). The point \(P(2, 0)\) is not on the graph because \(g(2)=4\ne 0\).

- What relationship does the line equation give between the vertex coordinates? - Substitute that relationship into vertex form. - What type of equation results after substituting the given point? - Check whether the parameter equation has more than one solution.

1. Since the vertex lies on \(y=x-2\), its coordinates satisfy \(k=h-2\). 2. Write \(g(x)=(x-h)^2+k\), and substitute \(k=h-2\): \(g(x)=(x-h)^2+h-2\). 3. Use \(P(4, 2)\): \(2=(4-h)^2+h-2\). 4. Expand and rearrange: \(2=h^2-7h+14\), so \(h^2-7h+12=0\). 5. Factor: \((h-3)(h-4)=0\). Thus, \(h=3\) or \(h=4\). 6. If \(h=3\), then \(k=1\), giving \(g_1(x)=(x-3)^2+1\). 7. If \(h=4\), then \(k=2\), giving \(g_2(x)=(x-4)^2+2\).

The two possible functions are \(g_1(x)=(x-3)^2+1\) \(g_2(x)=(x-4)^2+2\).

- Rewrite the family in vertex form to see how \(c\) affects the vertex. - What must the vertex's \(y\)-coordinate be when it lies on the \(x\)-axis? - What horizontal and vertical changes move the vertex from part b) to the origin?

1. Complete the square: \(p_c(x) = x^2 - 4x + c = (x - 2)^2 + c - 4\). The vertex is \((2, c - 4)\). For the vertex to lie on the \(x\)-axis, \(c - 4 = 0\), so \(c = 4\). 2. For \(c = 1\), \(p_1(x) = (x - 2)^2 - 3\), so the vertex is \((2, -3)\). 3. Solve \(x^2 - 4x + 1 = 0\): \(x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}\). 4. Moving \((2, -3)\) to \((0, 0)\) requires a translation \(2\) units left and \(3\) units up. The translated function is \(q(x) = x^2\).

a) \(c = 4\) b) Vertex \((2, -3)\); zeros \(x = 2 - \sqrt{3}\) and \(x = 2 + \sqrt{3}\) c) Translate \(2\) units left and \(3\) units up; \(q(x) = x^2\).

- Write each function in vertex form. - Use each vertex and opening direction to identify the function’s possible y-values. - For the algebraic check, set the functions equal and calculate the discriminant.

1. Complete the square: \(f(x) = (x - 2)^2 + 3\), so its vertex is \((2, 3)\). 2. Also, \(g(x) = -(x - 2)^2 - 1\), so its vertex is \((2, -1)\). 3. The graph of \(f\) opens upward and has minimum value \(3\), so \(f(x) \ge 3\). The graph of \(g\) opens downward and has maximum value \(-1\), so \(g(x) \le -1\). Therefore, the graphs cannot intersect. 4. Setting \(f(x) = g(x)\) gives \(2x^2 - 8x + 12 = 0\). 5. Its discriminant is \(D = (-8)^2 - 4 \cdot 2 \cdot 12 = 64 - 96 = -32\), confirming that there is no real intersection.

a) \(f\) has vertex \((2, 3)\), and \(g\) has vertex \((2, -1)\). b) Since \(f(x) \ge 3\) and \(g(x) \le -1\), the graphs cannot intersect. c) \(D = -32\), confirming no real intersection.

- Use the two zeros to write intercept form with an unknown leading coefficient. - The vertex x-coordinate is the midpoint of the zeros. - Use the vertex y-coordinate to determine the leading coefficient. - Average rate of change is the change in output divided by the change in input.

1. Begin with intercept form: \(f_a(x)=c(x-1)(x-a)\). The vertex lies halfway between the zeros, so its x-coordinate is \(\frac{a+1}{2}\). Substituting this value gives the vertex y-coordinate \(-\frac{c}{4}(a-1)^2\). Set this equal to \((a-1)^2\). Since \(a>1\), divide by \((a-1)^2\) to obtain \(c=-4\). Therefore, \(f_a(x)=-4(x-1)(x-a)\). 2. The vertex is \(\left(\frac{a+1}{2}, (a-1)^2\right)\), so vertex form is \(f_a(x)=-4\left(x-\frac{a+1}{2}\right)^2+(a-1)^2\). Expanding gives \(f_a(x)=-4x^2+(4a+4)x-4a\). 3. Since \(f_a(1)=0\) and \(f_a(2)=-4(2-1)(2-a)=4a-8\), the average rate of change is \(\frac{f_a(2)-f_a(1)}{2-1}=4a-8\).

1) \(f_a(x)=-4(x-1)(x-a)\) 2) Vertex form: \(f_a(x)=-4\left(x-\frac{a+1}{2}\right)^2+(a-1)^2\) Standard form: \(f_a(x)=-4x^2+(4a+4)x-4a\) 3) The average rate of change is \(4a-8\).

- Set two family members with different parameter values equal. - Complete the square to find the vertex. - Replace the parameter with the vertex x-coordinate. - Tangency to the x-axis occurs when the vertex y-coordinate is zero.

1. Set two family members equal: \(x^2-2ax+a=x^2-2bx+b\). For \(a\neq b\), this simplifies to \((a-b)(1-2x)=0\), so \(x=\frac{1}{2}\). Substitution gives \(y=\frac{1}{4}\). Thus \(P\left(\frac{1}{2}, \frac{1}{4}\right)\). 2. Complete the square: \(f_a(x)=(x-a)^2-a^2+a\). Therefore, the vertex is \((a, -a^2+a)\). 3. Since the vertex x-coordinate is \(x=a\), the locus is \(y=-x^2+x\). 4. A parabola is tangent to the x-axis when its vertex has y-coordinate zero. Solve \(-a^2+a=0\), giving \(a=0\) or \(a=1\). In the figure these are graphs q and r, respectively.

a) \(P\left(\frac{1}{2}, \frac{1}{4}\right)\) b) \((a, -a^2+a)\) c) \(y=-x^2+x\) d) Graph q has \(a=0\), and graph r has \(a=1\).

- Factor the leading coefficient from the first two terms. - Complete the square inside the parentheses. - Set the vertex form equal to zero and use the square-root property. - Intercept form has the structure \(a(x-r_1)(x-r_2)\).

1. Factor \(-0.5\) from the quadratic and linear terms: \(h(x)=-0.5(x^2-4x)+2.5\). 2. Complete the square: \(h(x)=-0.5\left((x-2)^2-4\right)+2.5=-0.5(x-2)^2+4.5\). 3. Set the vertex form equal to zero: \(-0.5(x-2)^2+4.5=0\). 4. Solve: \((x-2)^2=9\), so \(x-2=\pm3\). The zeros are \(x=-1\) and \(x=5\). 5. In context, the robot lands at \(x=5\,\text{in}\); the negative zero is outside the forward jump. 6. Using the zeros and leading coefficient, the intercept form is \(h(x)=-0.5(x+1)(x-5)\). 7. The graph shows the vertex \((2, 4.5)\) and x-intercepts \((-1, 0)\) and \((5, 0)\), confirming the results.

a) \(h(x)=-0.5(x-2)^2+4.5\) b) The zeros are \(x=-1\) and \(x=5\); the robot lands at \(5\,\text{in}\). c) \(h(x)=-0.5(x+1)(x-5)\) d) The graph confirms the vertex \((2, 4.5)\) and x-intercepts \((-1, 0)\) and \((5, 0)\).

- Substitute the coordinates of each common point. - Use opening direction, symmetry, vertices, and zeros to match the parabolas. - Substitute \(x=1\) and set the result equal to zero.

1. Substitution gives \(g_k(0)=2\) and \(g_k(-2)=4k-2(2k+1)+2=0\), independent of \(k\). 2. Graph p is \(g_{-1}(x)=-x^2-x+2\), which opens downward and has zeros \(-2\) and \(1\). 3. Graph q is \(g_{-0.5}(x)=-0.5x^2+2\), the downward-opening graph symmetric about the y-axis. 4. Graph r is \(g_{0.5}(x)=0.5(x+2)^2\), which is tangent to the x-axis at \((-2, 0)\). 5. Graph s is \(g_1(x)=x^2+3x+2\), with zeros \(-2\) and \(-1\). 6. For a zero at \(x=1\), \(g_k(1)=3k+3=0\), so \(k=-1\).

a) \(g_k(-2)=0\) and \(g_k(0)=2\) b) p: \(k=-1\); q: \(k=-0.5\); r: \(k=0.5\); s: \(k=1\) c) \(k=-1\)