Free math worksheets

- Compare the vertices of the two parabolas. - A horizontal shift changes the expression inside the square. - A vertical shift changes the constant outside the square.

1. The vertex of \(f\) is \((0, 0)\), while the vertex of \(g\) is \((-2, -2)\). 2. The vertex moves left \(2\) units and down \(2\) units. 3. A shift left \(2\) units replaces \(x\) with \(x+2\), and a shift down \(2\) units subtracts \(2\) outside the square. 4. Therefore, \(g(x)=(x+2)^2-2\).

Shift left \(2\) units and down \(2\) units; \(g(x)=(x+2)^2-2\)

- Compare the opening directions of the two parabolas. - Check whether corresponding y-values have opposite signs. - A reflection across the x-axis multiplies all outputs by \(-1\).

1. The two parabolas have the same width and the same axis of symmetry. 2. Every point \((x, y)\) on \(f\) corresponds to \((x, -y)\) on \(g\). For example, \(f(0)=-2\) and \(g(0)=2\). 3. This is a reflection across the x-axis, so \(g(x)=-f(x)\).

Reflect across the x-axis; \(g(x)=-f(x)\)

- Which coefficient can be read directly when \(x=0\)? - What does the sign of the coefficient of \(x^2\) tell you about the opening direction? - Use only information visible in the given standard form.

1. In standard form \(ax^2+bx+c\), the constant term gives \(f(0)\). Here \(f(0)=5\), so the y-intercept is \((0,5)\). 2. The leading coefficient is \(2>0\), so the parabola opens upward.

The y-intercept is \((0,5)\), and the parabola opens upward.

- Compare the expression with \(a(x-h)^2+k\). - Be careful with the sign inside the squared binomial. - Use the sign of the outside coefficient to decide maximum or minimum.

1. Vertex form is \(a(x-h)^2+k\). Here \(h=-2\) and \(k=7\), so the vertex is \((-2,7)\). 2. The axis of symmetry is \(x=-2\). 3. Since \(a=-3<0\), the parabola opens downward, so the vertex is a maximum.

Vertex: \((-2,7)\) Axis of symmetry: \(x=-2\) The vertex is a maximum.

- In factored form, when does the function equal zero? - Convert each zero into an ordered pair on the x-axis. - Check the sign of the overall scale factor.

1. The factors are zero when \(x=4\) or \(x=-1\), so the zeros are \(4\) and \(-1\). 2. The x-intercepts are \((4,0)\) and \((-1,0)\). 3. The leading coefficient is positive, so the parabola opens upward.

Zeros: \(x=-1\) and \(x=4\) x-intercepts: \((-1,0)\) and \((4,0)\) The parabola opens upward.

- How do vertical and horizontal shifts change a function rule? - For a shift to the right, decide whether the number inside the input should be added or subtracted. - Track which operations affect the entire function value and which affect the input.

1. A vertical stretch by a factor of \(3\) gives \(y = 3x^2\). 2. Shifting up \(1\) unit gives \(y = 3x^2 + 1\). 3. Shifting right \(2\) units replaces \(x\) with \(x-2\), so \(y = 3(x-2)^2 + 1\).

b) \(y = 3(x-2)^2 + 1\)

- A reflection across the x-axis changes every output to its opposite. - A reflection across the y-axis replaces each input \(x\) with \(-x\). - Use parentheses when substituting \(-x\), especially inside powers. - Simplify each resulting rule when possible.

1. A reflection across the x-axis is given by \(g(x)=-f(x)\). 2. For part a, \(g(x)=-(x^2-4x)=-x^2+4x\). 3. For part b, \(g(x)=-\left[\frac{1}{2}(x+2)^2-3\right]= -\frac{1}{2}(x+2)^2+3\). 4. A reflection across the y-axis is given by \(h(x)=f(-x)\). 5. For part a, \(h(x)=(-x)^2-4(-x)=x^2+4x\). 6. For part b, \(h(x)=\frac{1}{2}(-x+2)^2-3=\frac{1}{2}(x-2)^2-3\).

a) \(g(x)=-x^2+4x\); \(h(x)=x^2+4x\) b) \(g(x)=-\frac{1}{2}(x+2)^2+3\); \(h(x)=\frac{1}{2}(x-2)^2-3\)

- In vertex form, identify the horizontal and vertical shifts. - Replacing \(x\) with \(x-c\) moves the vertex horizontally. - A constant outside the squared expression moves the vertex vertically. - A vertical scale factor does not move a vertex that lies on the horizontal shift line.

1. For \(g(x)=(x+4)^2-2\), the graph shifts left \(4\) units and down \(2\) units, so the vertex is \((-4, -2)\). 2. For \(h(x)=3(x-5)^2+6\), the graph shifts right \(5\) units and up \(6\) units. The vertical stretch does not change the translated vertex, so the vertex is \((5, 6)\). 3. For \(k(x)=-x^2+4\), the graph reflects across the x-axis and shifts up \(4\) units. The vertex is \((0, 4)\).

a) Vertex: \((-4, -2)\); shift left \(4\) units and down \(2\) units b) Vertex: \((5, 6)\); shift right \(5\) units, vertically stretch by a factor of \(3\), and shift up \(6\) units c) Vertex: \((0, 4)\); reflect across the x-axis and shift up \(4\) units

- Decide whether each transformation changes inputs or outputs. - Track the coordinates of the given vertex. - A negative outside coefficient reflects the graph across the x-axis. - A reflection across the x-axis switches maxima and minima.

1. The rule \(g(x)=f(x)-8\) shifts the graph down \(8\) units. The maximum moves to \((2, -2)\). 2. The rule \(h(x)=f(x+3)\) shifts the graph left \(3\) units. The maximum moves to \((-1, 6)\). 3. The rule \(k(x)=-2f(x)\) vertically stretches by a factor of \(2\) and reflects across the x-axis. The point \((2, 6)\) maps to \((2, -12)\), and the maximum becomes a minimum.

a) Maximum at \((2, -2)\) b) Maximum at \((-1, 6)\) c) Minimum at \((2, -12)\)

- Recognize \(f(x)\) as a complete expression inside the rule for \(g\). - An outside factor changes all outputs. - A constant added outside the function creates a vertical shift. - Pay attention to the sign of the scale factor.

1. Since \(g(x)=-3f(x)+4\), multiply every output of \(f\) by \(-3\). 2. This reflects the graph across the x-axis and vertically stretches it by a factor of \(3\). 3. Then shift the result up \(4\) units.

Reflect across the x-axis, vertically stretch by a factor of \(3\), and then shift up \(4\) units.

- Use inputs equally far from the proposed vertical axis. - Substitute \(-5+h\) and \(-5-h\). - Compare the resulting squared expressions.

1. Symmetry about \(x=-5\) requires equal outputs at inputs equally far from \(-5\). 2. \(f(-5+h)=h^2-2\). 3. \(f(-5-h)=(-h)^2-2=h^2-2\). 4. Therefore, \(f(-5+h)=f(-5-h)\) for every real \(h\), proving symmetry about \(x=-5\).

Both \(f(-5+h)\) and \(f(-5-h)\) simplify to \(h^2-2\). Therefore, the graph is symmetric about \(x=-5\).

- Compare the vertices of the two parabolas. - Determine whether the transformed parabola opens in the same direction as the parent graph. - Use vertex form to combine the reflection and shifts.

1. The parent vertex is \((0, 0)\), and the graph shows the transformed vertex at \((2, 1)\). Therefore, the graph shifts right \(2\) units and up \(1\) unit. 2. The transformed parabola opens downward, so it is reflected across the x-axis. 3. Combining the transformations gives \(g(x)=-(x-2)^2+1\).

a) Reflect across the x-axis, shift right \(2\) units, and shift up \(1\) unit. b) \(g(x)=-(x-2)^2+1\)

- Read the vertex of each parabola. - Use vertex form \(y=a(x-h)^2+k\). - Determine whether each parabola opens upward or downward. - Compare the width with the parent parabola to determine \(|a|\).

1. The green parabola has vertex \((2, -3)\), opens upward, and has the same width as the parent graph. Therefore, \(g(x)=(x-2)^2-3\). 2. Expanding gives \(g(x)=x^2-4x+1\). 3. The purple parabola has vertex \((-3, 4)\), opens downward, and has the same width as the parent graph. Therefore, \(h(x)=-(x+3)^2+4\). 4. Expanding gives \(h(x)=-x^2-6x-5\).

\(g(x)=(x-2)^2-3\), or \(g(x)=x^2-4x+1\) \(h(x)=-(x+3)^2+4\), or \(h(x)=-x^2-6x-5\)

- Compare each transformed vertex with the vertex of \(f\). - Horizontal shifts change the expression inside the square. - Vertical shifts change the constant outside the square.

1. The vertex of \(f\) is \((0, 0)\). The vertex of \(g\) is \((2, 1)\), so \(g\) is shifted right \(2\) units and up \(1\) unit. 2. Therefore, \(g(x)=(x-2)^2+1\). 3. The vertex of \(h\) is \((-3, -1)\), so \(h\) is shifted left \(3\) units and down \(1\) unit. 4. Therefore, \(h(x)=(x+3)^2-1\).

\(g(x)=(x-2)^2+1\) \(h(x)=(x+3)^2-1\)

- Compare the vertices of the blue and red graphs. - Decide whether the change affects x-coordinates or y-coordinates. - A vertical shift preserves the shape and opening direction. - A reflection across the x-axis reverses the opening direction.

1. In panel a), the vertex moves from \((0, -1)\) to \((2, -1)\), so the graph shifts right \(2\) units. This matches \(g(x)=f(x-2)\). 2. In panel b), the vertex moves from \((0, -1)\) to \((0, 0.5)\), so the graph shifts up \(1.5\) units. This matches \(h(x)=f(x)+1.5\). 3. In panel c), the vertex moves from \((0, -1)\) to \((0, 1)\), and the parabola opens downward. This is a reflection across the x-axis and matches \(k(x)=-f(x)\).

a) \(g(x)=f(x-2)\), shift right \(2\) units; b) \(h(x)=f(x)+1.5\), shift up \(1.5\) units; c) \(k(x)=-f(x)\), reflect across the x-axis.

- Separate changes to the output from changes to the input. - Consider both the magnitude and sign of \(a\), using the stated restriction on \(a\). - Changes inside the input act in the opposite horizontal direction. - Use the transformed vertex to identify the example values.

1. Because \(a\ne0\), the parameter \(a\) scales all y-values by \(|a|\). If \(|a|>1\), the graph is stretched vertically; if \(0<|a|<1\), it is compressed vertically. If \(a<0\), the graph is also reflected across the x-axis. 2. The parameter \(c\) controls the horizontal shift. In \(f(x+c)\), a positive \(c\) shifts the graph left, while a negative \(c\) shifts it right. 3. The parameter \(d\) controls the vertical shift. A positive \(d\) shifts the graph up, while a negative \(d\) shifts it down. 4. In the graph, the parabola is reflected across the x-axis, shifted right \(2\) units, and shifted up \(1\) unit. Therefore, \(a=-1\), \(c=-2\), and \(d=1\).

For \(a\ne0\): \(a\): vertical scale by \(|a|\), with reflection across the x-axis when \(a<0\) \(c\): horizontal shift left for \(c>0\) and right for \(c<0\) \(d\): vertical shift up for \(d>0\) and down for \(d<0\) Shown example: \(a=-1\), \(c=-2\), \(d=1\)

- Locate the vertex of each parabola. - Measure the horizontal and vertical changes between the vertices. - A right shift uses a negative number inside the input.

1. The vertex of \(f\) is \((-1, 0)\), and the vertex of \(g\) is \((1, 1)\). 2. The vertex moves right \(2\) units and up \(1\) unit. 3. A shift right \(2\) units is represented by replacing \(x\) with \(x-2\), so \(c=-2\). 4. A shift up \(1\) unit gives \(d=1\). Therefore, \(g(x)=f(x-2)+1\).

\(c=-2\) and \(d=1\)

- Compare the vertices of the two parabolas. - Determine whether any vertical change occurred. - A shift right uses subtraction inside the input.

1. The vertex of \(f\) is \((-2, -2)\), and the vertex of \(g\) is \((3, -2)\). 2. The vertex moves right \(5\) units, with no vertical change. 3. A shift right \(5\) units replaces \(x\) with \(x-5\), so \(g(x)=f(x-5)\).

Shift right \(5\) units; \(g(x)=f(x-5)\)

- Compare the zeros of the two graphs. - Check that both zeros move by the same amount. - A shift left uses addition inside the input.

1. The zeros of \(f\) are \(x=0\) and \(x=2\). The corresponding zeros of \(g\) are \(x=-4\) and \(x=-2\). 2. Each zero moves left \(4\) units, so the entire graph shifts left \(4\) units. 3. A shift left \(4\) units replaces \(x\) with \(x+4\), so \(g(x)=f(x+4)\).

Shift left \(4\) units; \(g(x)=f(x+4)\)

- Compare the vertices of the two parabolas. - Check whether the shape or opening direction changed. - Horizontal shifts affect the input, and vertical shifts affect the output.

1. The vertex of \(f\) is \((0, -2)\), and the vertex of \(g\) is \((3, -1)\). 2. The vertex moves right \(3\) units and up \(1\) unit. 3. Therefore, \(g(x)=f(x-3)+1\).

Shift right \(3\) units and up \(1\) unit; \(g(x)=f(x-3)+1\)

- Compare the vertices relative to the y-axis. - Look for corresponding points with opposite x-coordinates. - A y-axis reflection replaces \(x\) with \(-x\).

1. The vertex of \(f\) is \((2, -1)\), while the vertex of \(g\) is \((-2, -1)\). 2. Corresponding points have opposite x-coordinates and equal y-coordinates. For example, the zero at \(x=3\) on \(f\) corresponds to the zero at \(x=-3\) on \(g\). 3. This is a reflection across the y-axis, which replaces \(x\) with \(-x\). 4. Therefore, \(g(x)=f(-x)\).

Reflect across the y-axis; \(g(x)=f(-x)\)

- How does the vertex determine the two shifts in the form \(a(x-h)^2+k\)? - Use the second point to determine the remaining coefficient. - The vertex tells you both the symmetry line and whether the extreme value is a maximum or minimum. - For x-intercepts, set the function value equal to \(0\) and solve exactly.

1. Use vertex form \(g(x)=a(x-2)^2+3\). 2. Since \(A=(0,-1)\) lies on the graph, \(-1=a(0-2)^2+3=4a+3\). Therefore, \(a=-1\), so \(g(x)=-(x-2)^2+3\). 3. The vertex is \((2,3)\), so the axis of symmetry is \(x=2\) and the maximum value is \(3\). 4. For the x-intercepts, set \(g(x)=0\): \(0=-(x-2)^2+3\), so \((x-2)^2=3\). 5. Thus, \(x=2-\sqrt{3}\) or \(x=2+\sqrt{3}\).

a) \(g(x)=-(x-2)^2+3\) b) Axis of symmetry: \(x=2\); maximum value: \(3\) c) x-intercepts: \((2-\sqrt{3},0)\) and \((2+\sqrt{3},0)\)

- Use the vertex \((-b, c)\) to determine the signs of \(b\) and \(c\). - Use the opening direction to determine the sign of \(a\). - Check each sign separately before choosing an option.

1. The vertex is \((-2, 3)\). 2. For \(g(x)=a(x+b)^2+c\), the vertex is \((-b, c)\). Since \(-b=-2\), \(b=2>0\). Since the vertex y-coordinate is \(3\), \(c=3>0\). 3. The parabola opens downward, so \(a<0\). 4. Therefore, the correct sign pattern is option 2.

2) \(a<0\), \(b>0\), \(c>0\)

- Which equivalent quadratic form makes a minimum or maximum visible immediately? - What number would make the first three terms a perfect-square trinomial? - After rewriting, use the smallest possible value of the squared expression.

1. Rewrite the quadratic as \(x^2-8x+16-16+11\). 2. Then \(f(x)=(x-4)^2-5\). 3. The squared term is at least \(0\), so the minimum value is \(-5\) when \(x=4\).

\(f(x)=(x-4)^2-5\). The minimum value is \(-5\) at \(x=4\).

- Which equivalent form makes zeros visible directly? - Look for two binomial factors whose product reproduces all three terms. - Set each factor equal to zero only after the factorization is complete.

1. Factor the quadratic: \(2x^2+x-6=(2x-3)(x+2)\). 2. The factors are zero when \(x=\frac{3}{2}\) or \(x=-2\). 3. Therefore, the x-intercepts are \(\left(\frac{3}{2},0\right)\) and \((-2,0)\).

\(f(x)=(2x-3)(x+2)\). The x-intercepts are \((-2,0)\) and \(\left(\frac{3}{2},0\right)\).

- Ask which feature is visible without additional algebra in each form. - Zeros come from inputs that make a product equal zero. - A squared expression shifted vertically makes an extremum easy to read. - The y-intercept occurs when \(x=0\).

1. Factored form shows the zeros directly: \(x=1\) and \(x=5\). 2. Vertex form shows the minimum directly: the minimum value is \(-4\) at \(x=3\). 3. Standard form shows \(f(0)=5\), so the y-intercept is \((0,5)\).

a) Factored form; zeros \(x=1\) and \(x=5\) b) Vertex form; minimum \(-4\) at \(x=3\) c) Standard form; y-intercept \((0,5)\)

- What feature of the rewritten quadratic will make the horizontal and vertical shifts visible? - Can you rewrite the quadratic so one squared binomial is visible? - After rewriting, compare each part with \(af(x+c)+d\), paying close attention to signs.

1. Complete the square: \(g(x)=-0.5(x^2-4x)+1\). 2. Since \(x^2-4x=(x-2)^2-4\), \(g(x)=-0.5[(x-2)^2-4]+1=-0.5(x-2)^2+3\). 3. Therefore, \(g(x)=-0.5f(x-2)+3\), so \(a=-0.5\), \(c=-2\), and \(d=3\). 4. Starting with the graph of \(f\), shift right \(2\) units, vertically compress by a factor of \(0.5\) and reflect across the x-axis, then shift up \(3\) units.

\(g(x)=-0.5f(x-2)+3\). Transformations: shift right \(2\) units; vertically compress by a factor of \(0.5\) and reflect across the x-axis; shift up \(3\) units.

- A reflection across the x-axis changes the sign of the entire output. - For a horizontal shift, think about how the input of the function must change. - Apply the transformations in the stated order. - Track the vertex after each transformation.

1. Reflecting across the x-axis multiplies all outputs by \(-1\): \(f_1(x)=-\left[\frac{1}{2}(x-2)^2-4\right]= -\frac{1}{2}(x-2)^2+4\). 2. Shifting left \(2\) units replaces \(x\) with \(x+2\): \(g(x)=-\frac{1}{2}[(x+2)-2]^2+4=-\frac{1}{2}x^2+4\). 3. The vertex form is \(g(x)=-\frac{1}{2}(x-0)^2+4\), which simplifies to \(g(x)=-\frac{1}{2}x^2+4\).

Vertex form: \(g(x)=-\frac{1}{2}(x-0)^2+4\) Standard form: \(g(x)=-\frac{1}{2}x^2+4\)

- Write the function rule after each individual transformation. - Decide whether the scale factor applies to only \(x^2\) or to the entire current output. - Multiplying and then subtracting generally differs from subtracting and then multiplying. - Compare each final rule with \(g(x)\).

1. Lina first obtains \(0.5x^2\), then shifts down \(3\) units: \(g_L(x)=0.5x^2-3\). Her description produces \(g\). 2. Felix first obtains \(x^2-3\), then compresses the entire output: \(g_F(x)=0.5(x^2-3)=0.5x^2-1.5\). 3. Since \(g_F(x)\ne g(x)\), Felix’s description does not produce \(g\).

Only Lina’s description produces \(g(x)=0.5x^2-3\). Felix’s sequence produces \(g_F(x)=0.5x^2-1.5\).

- Write a function rule after each transformation in each sequence. - When the stretch is second, it applies to the entire shifted output. - A constant vertical distance means the difference of the two functions is constant. - Simplify \(g(x)-h(x)\).

1. Stretch first, then shift: \(g(x)=3f(x)-2=3x^2-2\). 2. Shift first, then stretch the entire output: \(h(x)=3(f(x)-2)=3(x^2-2)=3x^2-6\). 3. Compute the difference: \(g(x)-h(x)=(3x^2-2)-(3x^2-6)=4\). 4. Since the difference is \(4\) for every real \(x\), the vertical distance between the graphs is constantly \(4\) units.

The vertical distance is \(4\) units because \(g(x)=3x^2-2\), \(h(x)=3x^2-6\), and \(g(x)-h(x)=4\) for every real \(x\).

- A reflection across the x-axis changes the sign of the entire output. - Track the vertex after each transformation. - When a reflection is performed after a shift, the shifted output is also reflected. - One differing characteristic point is enough to show that the graphs are different.

1. The vertex of \(f(x)=x^2-1\) is \((0, -1)\). 2. Sequence A gives \(-f(x)=-(x^2-1)=-x^2+1\), then \(g(x)=-x^2+5\). Its vertex is \((0, 5)\). 3. Sequence B gives \(f(x)+4=x^2+3\), then \(h(x)=-(x^2+3)=-x^2-3\). Its vertex is \((0, -3)\). 4. Since the vertices differ, the resulting functions and graphs differ.

The vertex of \(g\) is \((0, 5)\), and the vertex of \(h\) is \((0, -3)\). The different vertices show that changing the order changes the result.

- Write a function rule for each order. - A reflection across the x-axis multiplies the entire current output by \(-1\). - Identical functions must have equal outputs for every input. - Set the two final expressions equal and solve for \(c\).

1. Reflect first, then shift: \(g(x)=-f(x)+c=-x^2+c\). 2. Shift first, then reflect the entire output: \(h(x)=-(f(x)+c)=-x^2-c\). 3. For the results to be identical, set \(-x^2+c=-x^2-c\). 4. This gives \(c=-c\), so \(2c=0\) and \(c=0\). 5. When \(c=0\), both sequences produce \(-x^2\).

\(c=0\). Only a zero vertical shift makes the two transformation orders produce the same function.

- Read the transformations from the vertex-form parameters. - Track the original vertex through the horizontal and vertical shifts. - Determine how a negative outside coefficient changes the opening direction.

1. Shift the graph right \(3\) units, vertically stretch by a factor of \(2\), reflect across the x-axis, and shift up \(5\) units. 2. The vertex moves from \((0, 0)\) to \((3, 5)\). 3. The negative outside coefficient reflects the upward-opening parabola across the x-axis, so the transformed parabola opens downward and its vertex is a maximum.

1. Shift right \(3\) units, vertically stretch by a factor of \(2\), reflect across the x-axis, and shift up \(5\) units. 2. \((3, 5)\) 3. The reflection across the x-axis changes the minimum into a maximum.

- A horizontal shift moves the symmetry line and the zeros by the same amount. - A factor inside the input changes horizontal scale. - Determine when a product \(c\cdot f(x)\) can equal zero. - Points on the x-axis have output \(0\).

1. The rule \(p(x)=f(x-2)\) shifts the graph right \(2\) units. Its symmetry line becomes \(x=2\), and its zeros move to \(-4+2=-2\) and \(4+2=6\). 2. The rule \(q(x)=f(0.5x)\) horizontally stretches the graph by a factor of \(2\). It remains symmetric about the y-axis, and its zeros become \(-8\) and \(8\). 3. If \(c\ne 0\), then \(c\cdot f(x)=0\) exactly when \(f(x)=0\). Therefore, a nonzero vertical scale factor does not change the zeros.

a) Symmetric about \(x=2\); zeros at \(x=-2\) and \(x=6\) b) Symmetric about the y-axis; zeros at \(x=-8\) and \(x=8\) c) A nonzero vertical scale factor does not change the zeros.

- Rewrite the quadratic in vertex form. - Express both vertex coordinates in terms of \(k\). - Eliminate \(k\) and include the restriction implied by \(k>0\).

1. Rewrite the function in vertex form: \(f_k(x)=-\frac{2}{k}(x-2k)^2+8k+1\). 2. Thus, the vertex has coordinates \((2k,8k+1)\). 3. From \(x=2k\), obtain \(k=\frac{x}{2}\). Substitute into the y-coordinate: \(y=8\left(\frac{x}{2}\right)+1=4x+1\). 4. Because \(k>0\), the vertex satisfies \(x>0\). Therefore, the locus is \(y=4x+1\), for \(x>0\).

\(y=4x+1\), for \(x>0\)

- A reflection across the x-axis changes the sign of the entire function. - A horizontal shift changes the input. - Vertical shifts and scales act on the output. - Use parentheses around the entire expression before applying an outside scale factor.

1. Reflecting across the x-axis gives \(-f(x)=-(x^2-3x)=-x^2+3x\). 2. Shifting up \(2\) units gives \(g(x)=-x^2+3x+2\). 3. Shifting right \(4\) units gives \(f(x-4)=(x-4)^2-3(x-4)\). 4. Vertically compressing the entire output gives \(h(x)=0.5[(x-4)^2-3(x-4)]\).

a) \(g(x)=-x^2+3x+2\) b) \(h(x)=0.5[(x-4)^2-3(x-4)]\)

- Look for a complete replacement of \(x\) by \(x+3\). - Rewrite each rule so that \(p(x)\) appears as a single expression. - An outside factor changes vertical scale and may create a reflection. - A constant added outside the function creates a vertical shift.

1. Since \(q(x)=p(x+3)\), the graph of \(q\) is the graph of \(p\) shifted left \(3\) units. 2. Factor the first two terms of \(r\): \(r(x)=-2(x^2-5x)-3=-2p(x)-3\). 3. Therefore, the graph of \(r\) is obtained by vertically stretching the graph of \(p\) by a factor of \(2\), reflecting it across the x-axis, and shifting it down \(3\) units.

a) Shift left \(3\) units. b) Vertically stretch by a factor of \(2\), reflect across the x-axis, and shift down \(3\) units.

- Use the given point to solve for the unknown coefficient. - Evaluate the square before multiplying by the coefficient. - Compare \(k(x)\) directly with \(h(x)\) by writing \(k(x)=ch(x)\). - A negative output scaling reflects a graph across the x-axis.

1. Substitute \((2,4)\): \(4=a(2^2)=4a\), so \(a=1\). 2. Then \(h(-3)=(-3)^2=9\). 3. Substitute \((1,-2)\) into \(k(x)=bx^2\): \(-2=b(1)^2\), so \(b=-2\). 4. Since \(k(x)=-2x^2=-2h(x)\), reflect the graph of \(h\) across the x-axis and stretch it vertically by a factor of \(2\). Except for the origin, the graph of \(k\) lies in Quadrants III and IV.

a) \(a=1\) b) \(h(-3)=9\) c) \(k(x)=-2h(x)\). Reflect the graph of \(h\) across the x-axis and stretch it vertically by a factor of \(2\). Except for the origin, the graph lies in Quadrants III and IV.

- Represent the zeros as \(r\) and \(3r\). - Use the zeros to write the monic quadratic as a product of two linear factors. - Expand the factors and compare coefficients with the given quadratic.

1. Let the zeros be \(r\) and \(3r\). 2. Because the quadratic is monic, its factored form is \((x-r)(x-3r)\). 3. Expand: \((x-r)(x-3r)=x^2-4rx+3r^2\). 4. Compare the constant term with \(x^2+px+12\): \(3r^2=12\), so \(r^2=4\) and \(r=2\) or \(r=-2\). 5. Compare the coefficient of \(x\): \(p=-4r\). Thus \(p=-8\) when \(r=2\), and \(p=8\) when \(r=-2\).

\(p = -8\) or \(p = 8\)

- Use the graph of \(h\) to write the vertex coordinates in terms of \(a\). - Use the vertex coordinates to write the parabola in vertex form. - For part b, the vertex's y-coordinate must satisfy both equations that describe its location. - Use the condition \(a>0\) when choosing among algebraic solutions.

1. Because the vertex lies on \(h(x)=\frac{1}{x}\), its coordinates are \(\left(a,\frac{1}{a}\right)\), where \(a>0\). 2. A downward-opening parabola with leading coefficient \(-1\) and that vertex is \(g_a(x)=-(x-a)^2+\frac{1}{a}\). 3. If the vertex also lies on \(y=x-\frac{3}{2}\), then \(\frac{1}{a}=a-\frac{3}{2}\). 4. Multiply by \(2a\): \(2=2a^2-3a\), so \(2a^2-3a-2=0=(a-2)(2a+1)\). 5. The candidates are \(a=2\) and \(a=-\frac{1}{2}\). Since \(a>0\), \(a=2\). 6. Therefore, the corresponding parabola is \(g_2(x)=-(x-2)^2+\frac{1}{2}\).

a) \(g_a(x)=-(x-a)^2+\frac{1}{a}\), where \(a>0\) b) \(a=2\), so \(g_2(x)=-(x-2)^2+\frac{1}{2}\)

- Multiply each given output by the outside scale factor. - A negative scale factor also reflects the graph across the x-axis. - Use finite differences to identify the function type. - An outside factor distributes to every term of the function rule.

1. Multiply each output by \(2.5\): the values of \(g\) are \(-10, 5, 10, 5, -10, -35\). 2. Multiply each output by \(-0.5\): the values of \(h\) are \(2, -1, -2, -1, 2, 7\). 3. The graph of \(g\) is a vertical stretch of the graph of \(f\) by a factor of \(2.5\). The graph of \(h\) is a vertical compression by a factor of \(0.5\) followed by a reflection across the x-axis. 4. The second differences of the \(f(x)\)-values are constant at \(-4\), so let \(f(x)=ax^2+bx+c\). From \(f(0)=4\), \(c=4\). Since \(f(-1)=f(1)\), \(b=0\). From \(f(1)=2\), \(a+4=2\), so \(a=-2\). 5. Therefore, \(f(x)=-2x^2+4\), \(g(x)=2.5(-2x^2+4)=-5x^2+10\), and \(h(x)=-0.5(-2x^2+4)=x^2-2\).

a) \(g(x)\)-values: \(-10, 5, 10, 5, -10, -35\) \(h(x)\)-values: \(2, -1, -2, -1, 2, 7\) b) \(g\) is a vertical stretch by a factor of \(2.5\). \(h\) is a vertical compression by a factor of \(0.5\) and a reflection across the x-axis. c) \(f(x)=-2x^2+4\), \(g(x)=-5x^2+10\), and \(h(x)=x^2-2\)

- Use one table value and the corresponding value of \(p\) to solve for \(k\). - The sign and magnitude of an outside factor determine reflection and vertical scale. - A product is zero when one of its factors is zero. - Consider what happens to an output that is already \(0\).

1. Use \(x=0\): \(p(0)=-4\) and \(q(0)=8\), so \(8=k(-4)\). Therefore, \(k=-2\). 2. Multiplying by \(-2\) vertically stretches the graph by a factor of \(2\) and reflects it across the x-axis. 3. Solve \(p(x)=0\): \(x^2-4=0\), so \(x=-2\) or \(x=2\). 4. Since \(q(x)=-2p(x)\) and \(-2\ne0\), \(q(x)=0\) exactly when \(p(x)=0\). Thus, both functions have the same zeros. 5. In general, multiplying a function by a nonzero constant does not change the x-values where the function equals \(0\).

a) \(k=-2\) b) Vertically stretch by a factor of \(2\) and reflect across the x-axis. c) Both functions have zeros at \(x=-2\) and \(x=2\). Multiplying a function by a nonzero constant does not change its zeros.

- Raising the starting position adds the same amount to every height. - A vertical shift does not change the horizontal location of the maximum. - Track how \(f(x)+c\) changes the coordinates of a point.

1. Raising the nozzle by \(2.5\) feet adds \(2.5\) to every height: \(h_{\mathrm{new}}(d)=h(d)+2.5=-0.5d^2+2d+3.5\). 2. This is a vertical shift up \(2.5\) units. 3. A vertical shift does not change the horizontal coordinate of the vertex. The vertical coordinate increases by \(2.5\), so \((2, 3)\) moves to \((2, 5.5)\).

a) \(h_{\mathrm{new}}(d)=-0.5d^2+2d+3.5\) b) Shift the graph up \(2.5\) units. c) The new vertex is \((2, 5.5)\).

- The power of a linear expression expands to a polynomial. - Factor the coefficient of \(x\) from the entire input. - An input coefficient of \(0.2\) creates a horizontal stretch by its reciprocal. - Use the factored form to read the horizontal shift.

1. The expression \((0.2x-4)^2\) is the square of a linear polynomial, so its expansion is a polynomial of degree \(2\). 2. Factor the input: \(0.2x-4=0.2(x-20)\). Thus, \(h(x)=f(0.2(x-20))\). 3. Horizontally stretch the graph of \(f\) by a factor of \(5\). 4. Shift the result right \(20\) units.

Yes. \(h\) is a quadratic polynomial function. One sequence is a horizontal stretch by a factor of \(5\), followed by a shift right \(20\) units.

- Write the vertex coordinates in terms of \(k\). - Eliminate \(k\) between the x- and y-coordinates. - “Above \(y=2\)” means the vertex's y-coordinate is greater than \(2\). - Factor the resulting quadratic inequality.

1. The x-coordinate of the vertex is \(x_v=-\frac{-2k}{2}=k\). 2. Substitute \(x=k\): \(y_v=k^2-2k^2+3k=-k^2+3k\). Thus, the vertex is \((k, -k^2+3k)\). 3. Since \(x=k\), eliminate the parameter to obtain the locus \(y=-x^2+3x\). 4. The vertex is above \(y=2\) when \(-k^2+3k>2\). This is equivalent to \((k-1)(k-2)<0\), which holds for \(1<k<2\).

a) \(y=-x^2+3x\) b) \(1<k<2\)

- Evaluate the function at \(x=0\). - Use the vertex formula for a quadratic. - Eliminate the parameter \(a\) from the two vertex coordinates. - Use \(a>0\) to determine the relevant part of the locus.

1. Evaluate the function at zero: \(g_a(0)=0\). Therefore, every graph passes through \((0, 0)\). 2. The vertex x-coordinate is \(x_H=-\frac{\sqrt{a}}{2(-\frac{1}{2})}=\sqrt{a}\). 3. Substitute \(x=\sqrt{a}\): \(g_a(\sqrt{a})=a-\frac{1}{2}a=\frac{a}{2}\). Thus, \(H_a=\left(\sqrt{a}, \frac{a}{2}\right)\). 4. Since \(x=\sqrt{a}\), we have \(a=x^2\), with \(x>0\). Substitute into \(y=\frac{a}{2}\) to obtain \(y=\frac{1}{2}x^2\) for \(x>0\).

a) \(g_a(0)=0\), so every graph passes through \((0, 0)\). b) \(H_a=\left(\sqrt{a}, \frac{a}{2}\right)\) c) \(y=\frac{1}{2}x^2\) for \(x>0\)

- Compare the vertices in each panel. - Horizontal and vertical shifts determine the vertex-form structure. - In part b, use another readable point to find the vertical scale factor. - Substitute that point into \(a(x-h)^2+k\).

1. In part a, the parent vertex \((0, 0)\) moves to \((2, 1)\). The graph shifts right \(2\) units and up \(1\) unit, so \(g(x)=(x-2)^2+1\). 2. In part b, the parent vertex moves to \((-1, -3)\), giving the form \(g(x)=a(x+1)^2-3\). 3. The graph passes through \((0, -1)\), so \(-1=a(0+1)^2-3\). Thus, \(a=2\). 4. Therefore, \(g(x)=2(x+1)^2-3\), which is a vertical stretch by a factor of \(2\), a shift left \(1\) unit, and a shift down \(3\) units.

a) Shift right \(2\) units and up \(1\) unit; \(g(x)=(x-2)^2+1\). b) Vertically stretch by a factor of \(2\), shift left \(1\) unit, and shift down \(3\) units; \(g(x)=2(x+1)^2-3\).

- Begin with the vertex in each panel. - Write a vertex-form rule with an unknown outside coefficient. - Use a second readable point to determine that coefficient. - The sign and magnitude of the coefficient determine reflection and vertical scale.

1. In panel 1, the vertex is \((3, 4)\), so \(g(x)=a(x-3)^2+4\). 2. The point \((1, 2)\) lies on the graph. Substitution gives \(2=4a+4\), so \(a=-0.5\). Thus, \(g(x)=-0.5(x-3)^2+4\). 3. The transformations are a vertical compression by a factor of \(0.5\), a reflection across the x-axis, a shift right \(3\) units, and a shift up \(4\) units. 4. In panel 2, the vertex is \((-2, -1)\), so \(h(x)=a(x+2)^2-1\). 5. The point \((0, 7)\) lies on the graph. Substitution gives \(7=4a-1\), so \(a=2\). Thus, \(h(x)=2(x+2)^2-1\). 6. The transformations are a vertical stretch by a factor of \(2\), a shift left \(2\) units, and a shift down \(1\) unit.

a) Vertically compress by a factor of \(0.5\), reflect across the x-axis, shift right \(3\) units, and shift up \(4\) units; \(g(x)=-0.5(x-3)^2+4\). b) Vertically stretch by a factor of \(2\), shift left \(2\) units, and shift down \(1\) unit; \(h(x)=2(x+2)^2-1\).

- Begin with the vertex of the original quadratic. - A constant outside the function creates a vertical shift. - A change inside the input creates a horizontal shift in the opposite direction. - Match the translated vertices to the panels.

1. The graph of \(f\) has vertex \((0, -2)\), which appears in panel B. 2. The graph of \(g(x)=f(x)-2\) shifts down \(2\) units, so its vertex is \((0, -4)\). This is panel C. 3. The graph of \(h(x)=f(x+2)\) shifts left \(2\) units, so its vertex is \((-2, -2)\). This is panel D. 4. The graph of \(k(x)=f(x-2)\) shifts right \(2\) units, so its vertex is \((2, -2)\). This is panel A.

\(f\): B; \(g\): C; \(h\): D; \(k\): A

- Rewrite the quadratic using \(x-2\), or substitute both expressions directly. - Compare the even powers of \(a\). - Interpret \(2+a\) and \(2-a\) as inputs equally far from the proposed axis.

1. Rewrite the quadratic as \(f(x)=(x-2)^2+3\). 2. Substitute \(x=2+a\): \(f(2+a)=a^2+3\). 3. Substitute \(x=2-a\): \(f(2-a)=(-a)^2+3=a^2+3\). 4. Therefore, \(f(2+a)=f(2-a)\) for every real \(a\). 5. The inputs \(2+a\) and \(2-a\) are equally far from \(2\). Equal outputs at every such pair mean the graph is symmetric about \(x=2\).

a) \(f(2+a)=f(2-a)=a^2+3\) b) Points equally far to the left and right of \(x=2\) have equal y-values, so the graph is symmetric about \(x=2\).

- Compare the vertices first to find the horizontal and vertical shifts. - Compare opening directions to identify a reflection. - Compare widths to determine the magnitude of the vertical scale factor. - Check the parameters by writing each transformed rule.

1. The graph of \(f\) has vertex \((0, 2)\), opens downward, and has vertical scale factor \(1\). 2. The graph of \(g_1\) has vertex \((2, 1)\), opens downward, and has the same width. Thus, \(a=1\), \(c=2\), and \(d=-1\). 3. The graph of \(g_2\) has vertex \((-1, 0)\), opens upward, and has the same width. The reflection gives \(a=-1\), the left shift gives \(c=-1\), and the vertical placement gives \(d=2\), since \(-f(x+1)+2=(x+1)^2\).

\(g_1\): \(a=1\), \(c=2\), \(d=-1\) \(g_2\): \(a=-1\), \(c=-1\), \(d=2\)

- Compare vertices and opening directions. - Changes inside the input create horizontal transformations. - Changes outside the function create vertical transformations. - Eliminate rules that do not match the vertex or width.

1. In panel 1, the red vertex is \((1, -2)\), so the graph shifts right \(1\) unit and down \(2\) units. This is \(p_2(x)=f(x-1)-2\). 2. In panel 2, the red graph opens downward with vertex \((0, 1)\). This is a reflection across the x-axis followed by a shift up \(1\) unit, so it is \(p_4(x)=-f(x)+1\). 3. In panel 3, the red graph has the same vertex and opens upward but is narrower than the parent graph. Its outputs are doubled at the same x-values, so it is a vertical stretch by a factor of \(2\): \(p_1(x)=2f(x)\).

1: \(p_2(x)=f(x-1)-2\), shift right \(1\) unit and down \(2\) units; 2: \(p_4(x)=-f(x)+1\), reflect across the x-axis and shift up \(1\) unit; 3: \(p_1(x)=2f(x)\), vertically stretch by a factor of \(2\)

- Compare the vertices of the blue and red graphs. - Check whether each red parabola opens in the same direction as \(f\). - A change inside the input creates a horizontal transformation. - A change outside the function creates a vertical transformation. - Substitute the rule for \(f\) only after writing the transformation rule.

1. In panel a), the vertex moves from \((0, 0)\) to \((-1, 2)\), and the parabola opens downward. Therefore, reflect the graph across the x-axis, shift it left \(1\) unit, and shift it up \(2\) units. 2. Thus, \(g_1(x)=-f(x+1)+2=-(x+1)^2+2\). 3. In panel b), the vertex moves down to \((0, -1)\), and the graph is horizontally compressed by a factor of \(\frac{1}{2}\). Replacing \(x\) with \(2x\) produces that compression. 4. Thus, \(g_2(x)=f(2x)-1=(2x)^2-1=4x^2-1\).

a) \(g_1(x)=-f(x+1)+2=-(x+1)^2+2\) b) \(g_2(x)=f(2x)-1=4x^2-1\)

- Locate the vertex of each parabola. - Compare the horizontal and vertical positions of corresponding vertices. - Check whether the red graph opens upward or downward. - Horizontal shifts change the input, while vertical shifts change the output.

1. In panel a), the vertex moves from \((1, 0)\) to \((-2, -2)\), and the opening and width do not change. 2. This is a shift left \(3\) units and down \(2\) units, so \(g(x)=f(x+3)-2\). 3. In panel b), the vertex moves from \((1, 0)\) to \((1, 1)\), and the parabola changes from opening upward to opening downward. 4. This is a reflection across the x-axis followed by a shift up \(1\) unit, so \(g(x)=-f(x)+1\).

a) Shift left \(3\) units and down \(2\) units; \(g(x)=f(x+3)-2\) b) Reflect across the x-axis and shift up \(1\) unit; \(g(x)=-f(x)+1\)

- Compare the vertices of the two graphs. - A reflection across the y-axis replaces \(x\) with \(-x\). - A horizontal shift changes the input of the function. - Verify that your two rules simplify to the same quadratic expression.

1. The vertex of \(f\) is \((1, 0)\), and the vertex of \(g\) is \((-1, 1)\). 2. First method: Reflecting \(f\) across the y-axis moves its vertex to \((-1, 0)\). Shifting up \(1\) unit then gives \(g(x)=f(-x)+1\). 3. Second method: Shifting \(f\) left \(2\) units moves its vertex to \((-1, 0)\). Shifting up \(1\) unit then gives \(g(x)=f(x+2)+1\). 4. Because \(f(-x)+1=(-x-1)^2+1=(x+1)^2+1\) and \(f(x+2)+1=(x+1)^2+1\), both rules describe the same graph.

1) Reflect across the y-axis, then shift up \(1\) unit: \(g(x)=f(-x)+1\) 2) Shift left \(2\) units, then shift up \(1\) unit: \(g(x)=f(x+2)+1\)

- Identify the vertex of the blue graph first. - A translation preserves the width of a parabola. - A vertical stretch changes the width while keeping the same axis of symmetry. - Check one easy input after comparing the vertices.

1. The vertex of \(f\) is \((0, -1)\). 2. For \(p(x)=f(x-1)+2\), the graph shifts right \(1\) unit and up \(2\) units. Its vertex is \((1, 1)\), and it has the same width as \(f\), so \(p\) matches graph 1. 3. For \(q(x)=2f(x)-1\), the graph is stretched vertically by a factor of \(2\) and shifted down \(1\) unit. Its vertex is \((0, -3)\). 4. Also, \(f(1)=0\), so \(q(1)=2\cdot0-1=-1\). Graph 2 passes through \((1, -1)\), confirming the match.

Graph 1: \(p(x)=f(x-1)+2\); Graph 2: \(q(x)=2f(x)-1\)

- Compare the zeros of the blue and red graphs. - Use the movement of the zeros to determine the horizontal shift. - Compare the signs of corresponding y-values to determine \(a\).

1. The zeros of \(f\) are \(x=0\) and \(x=2\). The zeros of \(g\) are \(x=-1\) and \(x=1\). 2. Both zeros move left \(1\) unit, so the input is \(x+1\). Therefore, \(c=1\). 3. The vertex of \(f\) is \((1, -1)\). After shifting left \(1\) unit, it would be \((0, -1)\), but the red vertex is \((0, 1)\). 4. The y-values have been multiplied by \(-1\), so \(a=-1\). Thus, \(g(x)=-f(x+1)=-x^2+1\).

\(a=-1\) and \(c=1\)

- Compare the vertices to find the horizontal shift. - Compare the opening directions to identify a reflection. - Compare the widths and corresponding y-distances to determine the scale factor.

1. The vertex moves from \((0, 4)\) to \((1, -2)\), so the graph shifts right \(1\) unit. 2. The blue parabola opens upward, while the red parabola opens downward, so the graph is reflected across the x-axis. 3. The distance of the vertex from the x-axis changes from \(4\) to \(2\), and the red graph is wider. This corresponds to a vertical compression by a factor of \(\frac{1}{2}\). 4. Therefore, \(g(x)=-\frac{1}{2}f(x-1)=-\frac{1}{2}\left((x-1)^2+4\right)\).

\(g(x)=-\frac{1}{2}\left((x-1)^2+4\right)\)

- Compare the opening directions. - Compare the y-values of the two graphs at \(x=0\). - Separate the sign of the scale factor from its magnitude.

1. The blue graph opens upward, and the red graph opens downward, so the graph is reflected across the x-axis. 2. At \(x=0\), the graph shows \(f(0)=-4\) and \(g(0)=2\). Therefore, the output is multiplied by \(-\frac{1}{2}\), because \(-\frac{1}{2}(-4)=2\). 3. The factor \(\frac{1}{2}\) is a vertical compression, and the negative sign represents the reflection. 4. Thus, \(g(x)=-\frac{1}{2}f(x)\).

Reflect across the x-axis and vertically compress by a factor of \(\frac{1}{2}\); \(g(x)=-\frac{1}{2}f(x)\)

- Compare the vertices first. - Decide whether each red graph has the same width as the blue graph. - Horizontal shifts affect the input. - Vertical scaling and shifting affect the output.

1. In panel a), the vertex moves from \((0, 4)\) to \((-4, 3)\), and the width and opening direction remain unchanged. 2. Therefore, the graph shifts left \(4\) units and down \(1\) unit, so \(g(x)=f(x+4)-1\). 3. In panel b), the vertex moves from \((0, 4)\) to \((0, 3)\), and the red parabola is wider. 4. Multiplying outputs by \(0.5\) changes the vertex value from \(4\) to \(2\), and then shifting up \(1\) unit gives \(3\). Therefore, \(h(x)=0.5f(x)+1\).

a) Shift left \(4\) units and down \(1\) unit; \(g(x)=f(x+4)-1\) b) Vertically compress by a factor of \(0.5\) and shift up \(1\) unit; \(h(x)=0.5f(x)+1\)

- Begin by reading the two marked coordinates from the graph. - Substitute each common point into the general quadratic to determine how \(b\) and \(c\) depend on \(a\). - For part b, use the additional point only after the family equation is established. - Equal-height points on a parabola lie symmetrically about its axis. - Express the vertex y-coordinate in terms of \(a\) before considering the limit.

1. From the graph, \(P=(0,3)\) and \(Q=(4,3)\). 2. Since every graph passes through \(P\), \(c=3\). 3. Using \(Q\), \(3=16a+4b+3\). Thus, \(16a+4b=0\), so \(b=-4a\). Therefore, \(f_a(x)=ax^2-4ax+3\). 4. For \(R(2,5)\), \(5=4a-8a+3\). Thus, \(2=-4a\), so \(a=-0.5\). 5. The equal-height points \(P\) and \(Q\) are symmetric about \(x=2\), so the axis of symmetry is \(x=2\). 6. The vertex has y-coordinate \(f_a(2)=-4a+3\). As \(a\to\infty\), \(-4a+3\to-\infty\).

a) \(P=(0,3)\), \(Q=(4,3)\), and \(f_a(x)=ax^2-4ax+3\) b) \(a=-0.5\) c) The axis of symmetry is \(x=2\), and the vertex y-coordinate approaches \(-\infty\) as \(a\to\infty\).

- Use the axis of symmetry through the vertex to relate the two zeros. - After locating both zeros, include an unknown scale factor in factored form. - Use the vertex value to determine that scale factor. - Check that all three forms describe the same quadratic.

1. The axis of symmetry passes through the vertex, so it is \(x=2\). The zero at \(x=5\) is \(3\) units to the right of the axis, so the other zero is \(3\) units to the left: \(x=-1\). 2. Write \(f(x)=a(x+1)(x-5)\). At the vertex, \(f(2)=-9\), so \(-9=a(3)(-3)=-9a\). Thus, \(a=1\). 3. Factored form is \(f(x)=(x+1)(x-5)\). 4. Using the vertex gives \(f(x)=(x-2)^2-9\). 5. Expanding either form gives \(f(x)=x^2-4x-5\).

a) \(x=-1\) b) \(f(x)=(x+1)(x-5)\) c) Vertex form: \(f(x)=(x-2)^2-9\) Standard form: \(f(x)=x^2-4x-5\)

- Start with the x-intercepts, but remember that they do not determine the vertical scale by themselves. - Use one additional point from the graph to determine the scale factor. - The axis of symmetry lies halfway between the two zeros. - For the inequality, think about which form exposes the boundary values most directly.

1. The graph crosses the x-axis at \(x=-2\) and \(x=4\), so \(f(x)=a(x+2)(x-4)\). 2. The graph also shows the y-intercept \((0,-8)\). Substituting \(x=0\) gives \(-8=a(2)(-4)=-8a\), so \(a=1\). 3. Thus, \(f(x)=(x+2)(x-4)\). 4. The axis of symmetry is halfway between the zeros at \(x=1\), and the vertex is \((1,-9)\), so \(f(x)=(x-1)^2-9\). 5. Factored form makes the boundary zeros visible immediately, so together with the upward opening it shows \(f(x)<0\) for \(-2<x<4\).

a) \(f(x)=(x+2)(x-4)\) b) \(f(x)=(x-1)^2-9\) c) Factored form; \(f(x)<0\) for \(-2<x<4\) because the zeros are visible directly.

- Connect the coefficients of the quadratic with the condition required for its zeros to be real. - The vertex lies halfway between the zeros. - Expand \(a(x-x_1)(x-x_2)\) and compare the coefficient of \(x\).

1. Real zeros exist when the discriminant is nonnegative: \(D=k^2-4\cdot2\cdot8=k^2-64\ge0\). 2. Solving \(k^2\ge64\) gives \(k\le-8\) or \(k\ge8\). 3. The x-coordinate of the vertex is \(x=-\frac{k}{2\cdot2}=-\frac{k}{4}\). As a positive \(k\) becomes large, the midpoint of the zeros moves farther left, not closer to the origin. The claim is false. 4. If the zeros are \(x_1\) and \(x_2\), then \(g(x)=a(x-x_1)(x-x_2)\). 5. Expanding gives \(g(x)=ax^2-a(x_1+x_2)x+ax_1x_2\). Therefore, \(b=-a(x_1+x_2)\). 6. If \(x_1=-x_2\), then \(x_1+x_2=0\), so \(b=0\).

a) \(k\le-8\) or \(k\ge8\) b) The claim is false because the midpoint of the zeros is \(-\frac{k}{4}\), which moves away from the origin as positive \(k\) increases. c) Opposite zeros have sum \(0\), and coefficient comparison gives \(b=-a(x_1+x_2)=0\).

- A translation moves corresponding graph features by the same horizontal and vertical amounts. - Which feature of each parabola could help you determine the horizontal shift first? - After finding one shift, use another corresponding feature or the equations to determine the other. - Check your values by confirming that the transformed quadratic matches \(g\).

1. A shift right by \(c\) units and up by \(d\) units gives \(g(x)=f(x-c)+d\). 2. Substitute the rule for \(f\): \(g(x)=(x-c)^2-3(x-c)+d\). 3. Expand: \(g(x)=x^2+(-2c-3)x+(c^2+3c+d)\). 4. Compare the coefficient of \(x\): \(-2c-3=-9\), so \(c=3\). 5. Compare constants: \(c^2+3c+d=21\). Substituting \(c=3\) gives \(9+9+d=21\), so \(d=3\).

\(c=3\) and \(d=3\)

- Apply each transformation to the current function rule, one step at a time. - A vertical compression applied after a vertical shift also scales that shift. - Horizontal shifts commute with transformations that change only outputs.

1. Sequence 1 gives \((x+3)^2\), then \((x+3)^2-8\), then \(0.5[(x+3)^2-8]=0.5(x+3)^2-4\). It is correct. 2. Sequence 2 gives \((x+3)^2\), then \(0.5(x+3)^2\), then \(0.5(x+3)^2-8\). It is not correct. 3. Sequence 3 gives \(0.5x^2\), then \(0.5(x+3)^2\), then \(0.5(x+3)^2-4\). It is correct.

Sequences 1 and 3 are correct. Sequence 2 produces \(y=0.5(x+3)^2-8\), not \(g\).

- Complete the square to identify the vertex. - Compare the vertex's y-coordinate with zero. - Substitute the proposed common point into the family equation.

1. Complete the square: \(g_k(x)=\left(x-\frac{k}{2}\right)^2-\frac{k^2}{4}+2k-3\). Therefore, the vertex is \(\left(\frac{k}{2},-\frac{k^2}{4}+2k-3\right)\). 2. Because each parabola opens upward, the number of real zeros depends on the vertex's y-coordinate. Factor it: \(-\frac{k^2}{4}+2k-3=-\frac{1}{4}(k-2)(k-6)\). 3. The vertex is below the x-axis when \(k<2\) or \(k>6\), on the x-axis when \(k=2\) or \(k=6\), and above it when \(2<k<6\). Therefore, the graph has two, one, or no real zeros respectively. 4. Substitute \(x=2\): \(g_k(2)=4-2k+2k-3=1\). Thus, \((2,1)\) lies on every graph.

a) \(g_k(x)=\left(x-\frac{k}{2}\right)^2-\frac{k^2}{4}+2k-3\); vertex \(\left(\frac{k}{2},-\frac{k^2}{4}+2k-3\right)\) b) Two real zeros for \(k<2\) or \(k>6\); one real zero for \(k=2\) or \(k=6\); no real zeros for \(2<k<6\) c) \(g_k(2)=1\) for every \(k\).

- Complete the square to find the vertex. - Eliminate the parameter from the vertex coordinates. - Compare the vertex's y-coordinate with zero to count real zeros. - Collect the terms containing \(a\) to find where the function value is independent of the parameter.

1. Complete the square: \(f_a(x)=(x-a)^2-a^2+3a\). Therefore, the vertex is \((a,-a^2+3a)\). 2. At a vertex, \(x=a\). Replacing \(a\) with \(x\) in the y-coordinate gives the locus \(y=-x^2+3x\). 3. Each parabola opens upward, so the number of real zeros depends on \(-a^2+3a=a(3-a)\). It is negative for \(a<0\) or \(a>3\), zero for \(a=0\) or \(a=3\), and positive for \(0<a<3\). 4. Thus, there are two real zeros for \(a<0\) or \(a>3\), one real zero for \(a=0\) or \(a=3\), and no real zeros for \(0<a<3\). 5. Rewrite \(f_a(x)=x^2+a(3-2x)\). For the value to be independent of \(a\), require \(3-2x=0\), so \(x=\frac{3}{2}\). Then \(y=\frac{9}{4}\). The common point is \(\left(\frac{3}{2},\frac{9}{4}\right)\).

a) \(f_a(x)=(x-a)^2-a^2+3a\); vertex \((a,-a^2+3a)\) b) \(y=-x^2+3x\) c) Two real zeros for \(a<0\) or \(a>3\); one for \(a=0\) or \(a=3\); none for \(0<a<3\) d) \(\left(\frac{3}{2},\frac{9}{4}\right)\)

- Complete the square to identify the vertex. - Eliminate \(t\) from the vertex coordinates. - When does the vertex of a downward-opening parabola lie on the x-axis? - Collect the terms containing \(t\) to find a parameter-independent point.

1. Complete the square: \(h_t(x)=-\left(x-\frac{t}{2}\right)^2+\frac{(t-4)^2}{4}\). 2. Therefore, the vertex is \(\left(\frac{t}{2},\frac{(t-4)^2}{4}\right)\). 3. From \(x=\frac{t}{2}\), obtain \(t=2x\). Substitution gives the vertex locus \(y=(x-2)^2\). 4. A downward-opening parabola is tangent to the x-axis when its vertex lies on the axis. Thus, \(\frac{(t-4)^2}{4}=0\), so \(t=4\). 5. Rewrite \(h_t(x)=-x^2+4+t(x-2)\). At \(x=2\), the parameter term is zero and \(h_t(2)=0\). Therefore, every graph passes through \((2,0)\).

a) \(h_t(x)=-\left(x-\frac{t}{2}\right)^2+\frac{(t-4)^2}{4}\); vertex \(\left(\frac{t}{2},\frac{(t-4)^2}{4}\right)\) b) \(y=(x-2)^2\) c) \(t=4\) d) \((2,0)\)

- Use the vertex formula for a quadratic. - Express the vertex coordinates in terms of \(a\), then eliminate the parameter. - Determine when the parabola opens downward. - Account for the sign of \(a\) when solving the inequality for the vertex's height.

1. The vertex formula gives \(x_v=\frac{2}{a}\). 2. Evaluate the function at the vertex: \(y_v=f_a\left(\frac{2}{a}\right)=a-\frac{4}{a}\). 3. From \(x=\frac{2}{a}\), obtain \(a=\frac{2}{x}\). Substituting gives \(y=\frac{2}{x}-2x\), with \(x\ne0\). 4. The vertex is a maximum when the parabola opens downward, so \(a<0\). 5. It lies below the x-axis when \(a-\frac{4}{a}<0\). Since \(a<0\), multiplying by \(a\) reverses the inequality: \(a^2-4>0\). Combined with \(a<0\), this gives \(a<-2\).

a) \(y=\frac{2}{x}-2x\), for \(x\ne0\) b) \(a<-2\)

- Use the vertex formula for a quadratic. - Find a direct relation between the vertex coordinates that does not contain \(t\). - Minimize the square of the distance instead of the distance itself. - Can you rewrite the distance expression using one positive quantity that appears repeatedly?

1. The vertex x-coordinate is \(x_v=-\frac{-4}{2t}=\frac{2}{t}\). Substitution gives \(y_v=t\left(\frac{2}{t}\right)^2-4\left(\frac{2}{t}\right)=-\frac{4}{t}\). Thus, \(S_t=\left(\frac{2}{t}, -\frac{4}{t}\right)\). 2. Since \(y_v=-2x_v\) and \(t>0\) implies \(x_v>0\), the locus is \(y=-2x\) for \(x>0\). 3. The squared distance to \(Q\) is \(d^2(t)=\left(\frac{2}{t}\right)^2+\left(5-\frac{4}{t}\right)^2\). 4. Let \(z=\frac{1}{t}>0\). Then \(d^2=4z^2+(5-4z)^2=20z^2-40z+25\). This quadratic has its minimum at \(z=1\), so \(t=1\). 5. The closest vertex is \(S_1=(2, -4)\).

1. \(S_t=\left(\frac{2}{t}, -\frac{4}{t}\right)\) 2. \(y=-2x\) for \(x>0\) 3. \(t=1\); the closest vertex is \((2, -4)\).

- Use the marked point on the red graph to determine the coefficient in \(h(x)=ax^2\). - Compare \(h(x)\) with a constant multiple of \(f(x)\). - Rewrite the coefficient as a square to express \(h\) in the form \(f(cx)\).

1. The red graph passes through \((1,4)\). If \(h(x)=ax^2\), then \(4=a(1)^2\), so \(h(x)=4x^2\). 2. Since \(h(x)=4f(x)\), the graph is a vertical stretch of \(f\) by a factor of \(4\). 3. Also, \(h(x)=f(2x)=(2x)^2=4x^2\). Replacing \(x\) with \(2x\) produces a horizontal compression by a factor of \(\frac{1}{2}\).

\(h(x)=4x^2\) Vertical stretch factor: \(4\) Horizontal compression factor: \(\frac{1}{2}\)

- Read the coordinates of the vertices. - Find a relationship between each vertex's x- and y-coordinates. - Use vertex form and a visible point to determine the vertical stretch.

1. The vertices shown are \((-2, 2)\), \((0, 0)\), \((2, -2)\), and \((4, -4)\). They follow the pattern \((t, -t)\). 2. A parabola with vertex \((t, -t)\) has the form \(f_t(x)=a(x-t)^2-t\). 3. The graph with vertex \((0, 0)\) passes through \((1, 1)\), so \(a=1\). 4. Therefore, one possible family is \(f_t(x)=(x-t)^2-t\).

\(f_t(x)=(x-t)^2-t\)