- Compare the positions of the vertices. - A horizontal shift changes the expression inside the square. - A vertical shift changes the term outside the square.

1. The vertex of \(f\) is \((0, 0)\), while the vertex of \(g\) is \((-2, -2)\). 2. The shift is left \(2\) and down \(2\). 3. 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.

- Where is the lowest point of the curve? - Which vertex coordinate belongs inside the square, and which belongs outside?

1. The vertex is \((2,-3)\). 2. Vertex form is \(a(x-h)^2+k\), where \((h,k)\) is the vertex. 3. Therefore, the blanks are \(2\) and \(-3\).

\(2\), then \(-3\); \(f(x)=a(x-2)^2-3\)

- The two x-intercepts are the same distance from the axis of symmetry. - Find the midpoint of their x-coordinates.

1. The axis of symmetry lies halfway between the two zeros. 2. Their midpoint is \(\frac{-6+2}{2}=-2\). 3. The axis is \(x=-2\).

\(x=-2\)

- A vertical shift changes function outputs, not inputs. - Compare what the two rules do to the vertex \((0,0)\).

1. Shifting down changes every output by \(-5\). 2. Thus \(g(x)=f(x)-5=x^2-5\) is correct; \(h(x)\) shifts the graph right \(5\) units.

Use \(g(x)=x^2-5\), because the subtraction is outside the square and moves the graph down.

- 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.

- 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 \((2, 5.5)\).

- 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. - Check whether they open in the same direction. - Use vertex form to express the reflection and shifts.

1. The vertex moves from \((0, 0)\) to \((2, 1)\), a shift right \(2\) and up \(1\). 2. The transformed parabola opens downward, so reflect across the x-axis. 3. Therefore, \(g(x)=-(x-2)^2+1\).

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. - Decide whether each curve opens up or down. - Compare their widths with the parent parabola before choosing the coefficient.

1. The vertex of \(g\) is \((2, -3)\); it opens upward with the same width as \(f\). Thus \(g(x)=(x-2)^2-3\). 2. Expanding gives \(g(x)=x^2-4x+1\). 3. The vertex of \(h\) is \((-3, 4)\); it opens downward with the same width as \(f\). Thus \(h(x)=-(x+3)^2+4\). 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 vertex with the parent's vertex. - What expression inside a square becomes zero at each new vertex? - Substitute each vertex's x-coordinate into your rule to check its height.

1. The vertex of \(g\) is \((4, -1)\), so \(g(x)=(x-4)^2-1\). 2. The vertex of \(h\) is \((-3, -1)\), so \(h(x)=(x+3)^2-1\).

\(g(x)=(x-4)^2-1\) \(h(x)=(x+3)^2-1\)

- 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 both vertices. - Find their horizontal and vertical changes. - Test the sign of \(c\): at the new vertex's x-coordinate, does \(x+c\) equal the old vertex's x-coordinate?

1. The vertices are \((-1, 0)\) for \(f\) and \((1, 1)\) for \(g\). 2. The graph moves right \(2\), which gives \(c=-2\), and up \(1\), which gives \(d=1\). 3. Therefore, \(g(x)=f(x-2)+1\).

\(c=-2\), \(d=1\), and \(g(x)=f(x-2)+1\).

- Compare the vertices of the graphs. - Check whether either vertex changes height. - Write a shift to the right using subtraction in the input.

1. The vertex moves from \((-2,-2)\) to \((3,-2)\). 2. This is a shift right \(5\) units, with no vertical change. 3. Therefore, \(g(x)=f(x-5)\).

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 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 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; \(g(x)=f(-x)\)

- 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 \((0,5)\)

- Expand the squared binomial before combining constants. - Check the middle term; the two cross-products both matter.

1. Expand the square: \((x+3)^2=x^2+6x+9\). 2. Subtract \(4\): \(f(x)=x^2+6x+5\).

\(f(x)=x^2+6x+5\)

- Expand the two binomial factors first. - Then apply the coefficient outside the product to every term. - Check the sign of the leading coefficient in the result.

1. Multiply the binomials: \((x-3)(x+1)=x^2-2x-3\). 2. Multiply by \(-2\): \(f(x)=-2x^2+4x+6\).

\(f(x)=-2x^2+4x+6\)

- Each zero gives a factor that becomes zero at that input. - Use the y-intercept to find the remaining scale factor. - Substitute that value into the factored rule to check it.

1. The zeros give \(f(x)=a(x-2)(x-6)\). 2. Using the y-intercept: \(24=a(-2)(-6)=12a\), so \(a=2\). 3. Hence \(f(x)=2(x-2)(x-6)\).

\(f(x)=2(x-2)(x-6)\)

- What information does the vertex give you about the two shifts? - Use the other point to determine the coefficient outside the square. - Check both given points with your final rule.

1. With vertex \((1,-4)\), write \(f(x)=a(x-1)^2-4\). 2. Substitute \((3,4)\): \(4=a(3-1)^2-4\), so \(8=4a\) and \(a=2\). 3. Therefore, \(f(x)=2(x-1)^2-4\).

\(f(x)=2(x-1)^2-4\)

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

- 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 \(g\).

- 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.

- 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. - Use the vertex to find the horizontal and vertical shifts. - In part b, use another readable point to find the vertical scale factor.

1. In part a, the vertex moves from \((0, 0)\) to \((2, 1)\). This is a shift right \(2\) and up \(1\), so \(g(x)=(x-2)^2+1\). 2. In part b, the vertex is \((-1, -3)\), giving \(g(x)=a(x+1)^2-3\). 3. The point \((0, -1)\) gives \(-1=a-3\), so \(a=2\). 4. Thus \(g(x)=2(x+1)^2-3\): a vertical stretch by \(2\), a shift left \(1\), and a shift down \(3\).

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 \(x=2\).

- Use the vertex x-coordinate to find \(c\). - Compare opening directions and widths to find \(a\). - Substitute the transformed vertex into \(af(x-c)+d\) to find \(d\); a reflection also changes the parent vertex height.

1. The graph of \(f\) has vertex \((0, 2)\) and opens downward. 2. The vertex of \(g_1\) is \((2, 1)\). It opens downward with the same width, so \(a=1\), \(c=2\), and \(d=-1\), since \(f(x-2)-1=-(x-2)^2+1\). 3. The vertex of \(g_2\) is \((-1, 0)\). It opens upward with the same width, so \(a=-1\) and \(c=-1\). Because \(-f(x+1)+d=(x+1)^2-2+d\), the vertex height \(0\) gives \(d=2\).

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

- Compare the vertices and opening directions. - Changes inside the input move the graph horizontally. - Compare heights at the same x-value to distinguish vertical scaling from horizontal scaling.

1. In panel 1, the vertex moves to \((1, -2)\); this is \(p_2(x)=f(x-1)-2\), a shift right \(1\) and down \(2\). 2. In panel 2, the parabola opens downward with vertex \((0, 1)\); this is \(p_4(x)=-f(x)+1\), a reflection across the x-axis followed by a shift up \(1\). 3. In panel 3, the vertex stays fixed, but the parabola is narrower and its outputs double; this is \(p_1(x)=2f(x)\), a vertical stretch by \(2\).

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 \(2\).

- Read the vertex and opening direction in each panel. - In part a, apply the reflection to the whole parent output. - In part b, use the width to determine the positive value of \(b\). - Expand after writing each transformation rule.

1. In part a, the vertex moves to \((-1, 2)\) and the parabola opens downward. Reflect across the x-axis, shift left \(1\), and shift up \(2\). 2. Thus \(g_1(x)=-f(x+1)+2=-(x+1)^2+2=-x^2-2x+1\). 3. In part b, the vertex is \((0, -1)\). The parabola is horizontally compressed by a factor of \(\frac{1}{2}\). Since \(b>0\), the input coefficient is \(b=2\). 4. Thus \(g_2(x)=f(2x)-1=4x^2-1\). Without the condition \(b>0\), \(b=-2\) would also give the same graph.

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

- Locate the vertex of each parabola. - Compare the movement of the vertices. - Check whether the transformed graph opens up or down. - Write the input and output changes separately.

1. In panel a), the vertex moves from \((1, 0)\) to \((-2, -2)\), and the width and opening stay the same. 2. Shifting left \(3\) units and down \(2\) units gives \(g(x)=f(x+3)-2\). 3. In panel b), the vertex moves from \((1, 0)\) to \((1, 1)\) while the opening reverses. 4. Reflecting across the x-axis and then shifting up \(1\) unit gives \(g(x)=-f(x)+1\).

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 two vertices. - A reflection across the y-axis replaces every input \(x\) by \(-x\). - Check whether a horizontal shift could place the vertex in the same spot. - Simplify the two rules to check that they agree.

1. The vertex moves from \((1, 0)\) to \((-1, 1)\). 2. Reflect across the y-axis to reach \((-1, 0)\), then shift up \(1\): \(g(x)=f(-x)+1\). 3. Alternatively, shift left \(2\) and then up \(1\): \(g(x)=f(x+2)+1\). 4. Both give \((x+1)^2+1\), since \((-x-1)^2=(x+1)^2\).

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: \(g(x)=f(x+2)+1\)

- Use the marked point to find the coefficient in \(h(x)=ax^2\). - Compare \(h(x)\) with a multiple of \(f(x)\). - What happens to \(x^2\) when the input \(x\) is multiplied by a number?

1. The marked point \((1, 4)\) is on \(h\). Writing \(h(x)=ax^2\) gives \(4=a\), so \(h(x)=4x^2\). 2. Since \(h(x)=4f(x)\), the vertical stretch factor is \(4\). 3. Also \(h(x)=f(2x)=(2x)^2=4x^2\). This is a horizontal compression by a factor of \(\frac{1}{2}\).

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

- Find the parent vertex first. - A translation does not change the width. - A vertical stretch changes the widths and the vertex height. - Check one simple input after comparing the vertices.

1. The vertex of \(f\) is \((0, -1)\). 2. The graph of \(p\) shifts right \(1\) and up \(2\) to vertex \((1, 1)\), keeping the same width. It matches curve \(u\). 3. The graph of \(q\) stretches by \(2\), then shifts down \(1\), moving its vertex to \((0, -3)\). 4. Since \(f(1)=0\), we have \(q(1)=-1\), another point on curve \(v\).

Curve \(u\) is \(p\): vertex \((1, 1)\), with the same width as \(f\). Curve \(v\) is \(q\): vertex \((0, -3)\), narrower than \(f\) because of the vertical stretch.

- Compare the pairs of x-intercepts. - Track the direction of the horizontal shift. - Compare corresponding vertex heights to determine the scale and reflection.

1. The zeros of \(f\) are \(0\) and \(2\), while the zeros of \(g\) are \(-1\) and \(1\). 2. Both zeros move left \(1\), so the input is \(x+1\), giving \(c=1\). 3. The parent vertex \((1,-1)\) would move to \((0,-1)\). The new vertex is \((0,1)\), so outputs are multiplied by \(-1\), giving \(a=-1\). 4. Therefore, \(g(x)=-f(x+1)=-x^2+1\).

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

- Compare the x-coordinates of the vertices to find \(c\). - The new opening direction tells you the sign of \(a\). - Use the vertex y-values to find \(a\); there is no extra vertical shift in the given form.

1. The vertex moves from \((0,4)\) to \((1,-2)\), so the horizontal shift gives \(c=1\). 2. The graph now opens downward, so \(a<0\). It is wider than \(f\), indicating a compression. 3. At the vertex, \(g(1)=a f(0)=4a=-2\), so \(a=-\frac{1}{2}\). 4. Therefore, \(g(x)=-\frac{1}{2}f(x-1)=-\frac{1}{2}((x-1)^2+4)\).

\(a=-\frac{1}{2}\), \(c=1\), and \(g(x)=-\frac{1}{2}((x-1)^2+4)\)

- 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; \(h(x)=0.5f(x)+1\)

- 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 \((2+\sqrt{3},0)\)

- Read the two marked points and substitute their coordinates. - Use each common point to find how \(b\) and \(c\) depend on \(a\). - For part b, use the extra point only after writing the family rule. - The axis lies halfway between two x-coordinates with equal y-values. - Write the vertex height as an expression in \(a\) and see what happens when positive \(a\) grows.

1. The marked points are \(P=(0,3)\) and \(Q=(4,3)\). 2. Substituting \(P\) gives \(c=3\); substituting \(Q\) gives \(3=16a+4b+3\), so \(b=-4a\). 3. Hence \(f_a(x)=ax^2-4ax+3\). 4. Substituting \(R(2,5)\) gives \(5=4a-8a+3\), so \(a=-\frac{1}{2}\). 5. The equal-height points are symmetric about \(x=2\); the vertex height is \(f_a(2)=3-4a\). 6. As positive \(a\) increases, \(3-4a\) decreases without bound, so the vertex moves farther and farther down.

a) \(P=(0,3)\), \(Q=(4,3)\), and \(f_a(x)=ax^2-4ax+3\) b) \(a=-\frac{1}{2}\) c) The axis is \(x=2\); the vertex height decreases without bound as positive \(a\) increases.

- 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.

- Start in the form that shows the given vertex directly. - The y-intercept can determine the remaining coefficient. - Which form is easiest to use when setting the function equal to zero?

1. The vertex gives \(f(x)=a(x-1)^2-9\). 2. Substitute the y-intercept: \(-5=a-9\), so \(a=4\). 3. Expand \(4(x-1)^2-9\) to get \(f(x)=4x^2-8x-5\). 4. Solve \(4(x-1)^2-9=0\): \((x-1)^2=\frac{9}{4}\), so \(x=1\pm\frac{3}{2}\). 5. The zeros are \(-\frac{1}{2}\) and \(\frac{5}{2}\).

a) \(f(x)=4x^2-8x-5\) b) \(x=-\frac{1}{2}\) and \(x=\frac{5}{2}\)

- 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 \(b=-a(x_1+x_2)=0\).

- 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 \((2, -4)\).

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

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