### Learning Outcomes

- Graph vertical and horizontal shifts of quadratic functions
- Graph vertical compressions andstretches of quadratic functions
- Write the equation of a transformed quadratic function using the vertex form
- Identify the vertex and axis of symmetry for a given quadratic function in vertex form

The **standard form of a quadratic function** presents the function in the form

[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]

where [latex]\left(h,\text{ }k\right)[/latex] is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the **vertex form of a quadratic function**.

## Shift Up and Down by Changing the Value of [latex]k[/latex]

You can represent a vertical (up, down) shift of the graph of [latex]f(x)=x^2[/latex] by adding or subtracting a constant, [latex]k[/latex].

[latex]f(x)=x^2 + k[/latex]

If [latex]k>0[/latex], the graph shifts upward, whereas if [latex]k<0[/latex], the graph shifts downward.### Example

Using an online graphing calculator, plot the function [latex]f(x)=x^2+k[/latex].Now change the [latex]k[/latex] value to shift the graph down 4 units, then up 4 units.

Answer:The equation for the graph of[latex]f(x)=x^2[/latex] that has been shifted up 4 units is[latex-display]f(x)=x^2+4[/latex-display]The equation for the graph of[latex]f(x)=x^2[/latex] that has been shifted down 4 units is[latex-display]f(x)=x^2-4[/latex-display]

## Shift left and right by changing the value of [latex]h[/latex]

You can represent a horizontal (left, right) shift of the graph of [latex]f(x)=x^2[/latex] by adding or subtracting a constant, [latex]h[/latex], to the variable [latex]x[/latex], before squaring.

[latex]f(x)=(x-h)^2 [/latex]

### tip for success

Remember that the negative sign inside the argument of the vertex form of a parabola (in the parentheses with the variable [latex]x[/latex] ) is part of the formula [latex]f(x)=(x-h)^2 +k[/latex].If [latex]h>0[/latex], we have[latex]f(x)=(x-h)^2 +k[/latex]. You'll see the negative sign, but the graph will shift right.If [latex]h<0[/latex], we have[latex]f(x)=(x-(-h))^2 +k \rightarrow f(x)=(x+h)^2+k[/latex]. You'll see the positive sign, but the graph will shift left.

### Example

Using an online graphing calculator, plot the function [latex]f(x)=(x-h)^2[/latex]. Now change the [latex]h[/latex] value to shift the graph 2 units to the right, then 2 units to the left.

Answer:The equation for the graph of[latex]f(x)=x^2[/latex] that has been shifted right 2 units is

[latex]f(x)=(x-2)^2[/latex]

The equation for the graph of[latex]f(x)=^2[/latex] that has been shifted left 2 units is

[latex]f(x)=(x+2)^2[/latex]

## Stretch or compress by changing the value of [latex]a[/latex].

You can represent a stretch or compression (narrowing, widening)of the graph of [latex]f(x)=x^2[/latex] bymultiplying the squared variable by a constant, [latex]a[/latex].

[latex]f(x)=ax^2 [/latex]

The magnitude of [latex]a[/latex]indicates the stretch of the graph. If [latex]|a|>1[/latex], the point associated with a particular [latex]x[/latex]-value shifts farther from the [latex]x[/latex]*-*axis, so the graph appears to become narrower, and there is a vertical stretch. But if [latex]|a|<1[/latex], the point associated with a particular [latex]x[/latex]-value shifts closer to the [latex]x[/latex]

*-*axis, so the graph appears to become wider, but in fact there is a vertical compression.

### Example

Using an online graphing calculator plot the function[latex]f(x)=ax^2[/latex]. Now adjust the [latex]a[/latex] value to create a graph that has been compressed vertically by a factor of [latex]\frac{1}{2}[/latex] and another that has been vertically stretched by a factor of 3. What are the equations of the two graphs?

Answer:The equation for the graph of[latex]f(x)=x^2[/latex] that has been compressed vertically by a factor of [latex]\frac{1}{2}[/latex] is

[latex]f(x)=\frac{1}{2}x^2[/latex]

The equation for the graph of[latex]f(x)=x^2[/latex] that has been vertically stretched by a factor of 3 is

[latex]f(x)=3x^2[/latex]

[latex]\begin{align}&a{\left(x-h\right)}^{2}+k=a{x}^{2}+bx+c\\ &a{x}^{2}-2ahx+\left(a{h}^{2}+k\right)=a{x}^{2}+bx+c \end{align}[/latex]

For the two sides to be equal, the corresponding coefficients must be equal. In particular, the coefficients of [latex]x[/latex] must be equal.

[latex]-2ah=b,\text{ so }h=-\dfrac{b}{2a}[/latex].

This is the [latex]x[/latex] coordinate of the vertexr and [latex]x=-\dfrac{b}{2a}[/latex] is the**axis of symmetry** we defined earlier. Setting the constant terms equal gives us:

[latex]\begin{align}a{h}^{2}+k&=c \\[2mm] k&=c-a{h}^{2} \\ &=c-a-{\left(\dfrac{b}{2a}\right)}^{2} \\ &=c-\dfrac{{b}^{2}}{4a} \end{align}[/latex]

In practice, though, it is usually easier to remember that [latex]h[/latex]is the output value of the function when the input is [latex]h[/latex], so [latex]f\left(h\right)=f\left(-\dfrac{b}{2a}\right)=k[/latex].### Example

Using an online graphing calculator plot the function [latex]f\left(x\right)\ =\ a\left(x-h\right)^2+k[/latex]. Now adjust the variables [latex]a,h,k[/latex] to define two quadratic functions whose axis of symmetry is [latex]x=-3[/latex], and whose vertex is [latex](-3, 2)[/latex].How many potential values are there for [latex]h[/latex] in this scenario? How about [latex]k[/latex]? How about [latex]a[/latex]

Answer:There is only one [latex](h,k)[/latex] pair that will satisfy these conditions,[latex](-3,2)[/latex]. The value of [latex]a[/latex] does not affect the line of symmetry or the vertex of a quadratic graph, so [latex]a[/latex] can take on any value.

## Challenge Problem

Define a function whose axis of symmetry is [latex]x = -3[/latex], and whose vertex is [latex](-3,2)[/latex] and has an average rate of change of 2 on the interval [latex][-2,0][/latex].Use the online graphing calculator where you plotted [latex]f\left(x\right)\ =\ a\left(x-h\right)^2+k[/latex] to help with this.### Try It

A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?

(credit: modification of work by Dan Meyer)

Answer:The path passes through the origin and has vertex at [latex]\left(-4,\text{ }7\right)[/latex], so [latex]\left(h\right)x=-\frac{7}{16}{\left(x+4\right)}^{2}+7[/latex]. To make the shot, [latex]h\left(-7.5\right)[/latex] would need to be about 4 but [latex]h\left(-7.5\right)\approx 1.64[/latex]; he doesn’t make it.

## Licenses & Attributions

### CC licensed content, Original

- Interactive: Transform Quadratic 1.
**Provided by:**Lumen Learning (with Desmos)**Located at:**https://www.desmos.com/calculator/fpatj6tbcn.**License:**CC BY: Attribution. - Interactive: Transform Quadratic 2.
**Provided by:**Lumen Learning (with Desmos)**Located at:**https://www.desmos.com/calculator/5g3xfhkklq.**License:**CC BY: Attribution. - Interactive: Transform Quadratic 3.
**Provided by:**Lumen Learning (with Desmos)**Located at:**https://www.desmos.com/calculator/ha6gh59rq7.**License:**CC BY: Attribution. - Interactive: Transform Quadratic 4.
**Provided by:**Lumen Learning (with Desmos)**Located at:**https://www.desmos.com/calculator/pimelalx4i.**License:**CC BY: Attribution. - Revision and Adaptation.
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### CC licensed content, Shared previously

- College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**Located at:**https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites.**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[emailprotected].