Vertical Shifts

If we shift the graph of the function [latex]f(x)=x^2[/latex] up 5 units, all of the points on the graph increase their [latex]y[/latex]-coordinates by 5, but their [latex]x[/latex]-coordinates remain the same. Therefore, the equation of the function [latex]f(x)=x^2[/latex] after it has been shifted up 5 units transforms to [latex]f(x)=x^2+5[/latex]. Table 1 shows the changes to specific values of this function, which are replicated on the graph in figure 1.

[latex]x[/latex]	[latex]x^2[/latex]	[latex]x^2+5[/latex]	Figure 1. Shifting the graph of [latex]f(x)=x^2[/latex] up 5 units. Every point [latex](x, y)[/latex] moves to ([latex](x, y+5)[/latex].
-3	9	14
-2	4	9
-1	1	6
0	0	5
1	1	6
2	4	9
3	9	14
Table 1. [latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=x^2+5[/latex]

If we shift the graph of the function [latex]f(x)=x^2[/latex] down 3 units, all of the points on the graph decrease their [latex]y[/latex]-coordinates by 3, but their [latex]x[/latex]-coordinates remain the same. Therefore, the equation of the function [latex]f(x)=x^2[/latex] after it has been shifted down 3 units transforms to [latex]f(x)=x^2-3[/latex]. Table 2 shows the changes to specific values of this function, which are replicated on the graph in figure 2.

[latex]x[/latex]	[latex]x^2[/latex]	[latex]x^2-3[/latex]	Figure 2. Shifting the graph of [latex]f(x)=x^2[/latex] down 3 units. Every point [latex](x, y)[/latex] moves to [latex](x, y-3)[/latex].
-3	9	6
-2	4	1
-1	1	-2
0	0	-3
1	1	-2
2	4	1
3	9	6
Table 2. [latex]f(x)=x^2[/latex] is transformed to [latex]f(x)=x^2-3[/latex]

 

Change the value of [latex]k[/latex] in the interactive desmos graph in figure 3 by moving the vertex up or down.

Figure 3. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=x^2+k[/latex]

Every point [latex](x, y)[/latex] moves to [latex](x, y+k)[/latex].

Horizontal Shifts

 

Change the value of [latex]h[/latex] in the interactive desmos graph in figure 6 by moving the vertex left or right.

Figure 6. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=(x-h)^2[/latex]

Stretching Up and Compressing Down

If we vertically stretch the graph of the function [latex]f(x)=x^2[/latex] by a factor of 2, all of the [latex]y[/latex]-coordinates of the points on the graph are multiplied by 2, but their [latex]x[/latex]-coordinates remain the same. Since the vertex of [latex]f(x)=x^2[/latex] is (0, 0), it transforms to (0, 0). That is, [latex]2\times0^2=2\times0=0[/latex]. The vertex, since it is an [latex]x[/latex]-intercept, stays at the same location (0, 0). In other words, the vertex is the anchor point in the stretching process. The equation of the function after the graph is stretched by a factor of 2 is [latex]f(x)=2x^2[/latex]. The reason for multiplying  [latex]x^2[/latex] by 2 is that each [latex]y[/latex]-coordinate is doubled, and since[latex]y=x^2[/latex], [latex]x^2[/latex] is doubled. Table 5 shows this change and the graph is shown in figure 7.​

[latex]x[/latex]	[latex]x^2[/latex]	[latex]2x^2[/latex]	Figure 7. [latex]f(x)=x^2[/latex] transformed to [latex]f(x)=2x^2[/latex]. Each [latex]y[/latex]-value is doubled.
-3	9	18
-2	4	8
-1	1	2
0	0	0
1	1	2
2	4	8
3	9	18
Table 5. Stretching the graph vertically by a factor of 2 transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=2x^2[/latex]

On the other hand, if our factor is [latex]\dfrac{1}{2}[/latex] and we vertically compress the graph of the function [latex]f(x)=x^2[/latex] all of the [latex]y[/latex]-coordinates of the points on the graph are halved, but their [latex]x[/latex]-coordinates remain the same. This means the [latex]y[/latex]-coordinates are multiplied by [latex]\dfrac{1}{2}[/latex], or divided by 2. The [latex]x[/latex]-intercept (0, 0) of [latex]f(x)=x^2[/latex] is the only point that doesn’t move. In other words, the [latex]x[/latex]-intercept is the anchor point in the compressing process. The equation of the function after being compressed is [latex]f(x)=\dfrac{1}{2}x^2[/latex]. The reason for multiplying [latex]x^2[/latex] by [latex]\dfrac{1}{2}[/latex] is that each [latex]y[/latex]-coordinate becomes half of the original value when it is divided by 2. Table 6 shows this change and the graph is shown in figure 8.

[latex]x[/latex]	[latex]x^2[/latex]	[latex]\frac{1}{2}x^2[/latex]	Figure 8. Compressing the graph when the factor is [latex]\frac{1}{2}[/latex]. Each [latex]y[/latex]-value is divided by 2.
-3	9	4.5
-2	4	2
-1	1	0.5
0	0	0
1	1	0.5
2	4	2
3	9	4.5
Table 6. When the factor is [latex]\dfrac{1}{2}[/latex] the graph is compressed and transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=\dfrac{1}{2}x^2[/latex]

Reflection across the [latex]x[/latex]-axis

When the graph of the function [latex]f(x)=x^2[/latex] is reflected across the [latex]x[/latex]-axis, the [latex]y[/latex]-coordinates of all of the points on the graph change their signs, from positive to negative values or from negative to positive values, while the [latex]x[/latex]-coordinates remain the same. The equation of the function after [latex]f(x)=x^2[/latex] is reflected across the [latex]x[/latex]-axis is [latex]f(x)=-x^2[/latex]. The graph changes from opening upwards to opening downwards. Table 7 shows the effect of such a reflection on the functions values and the graph is shown in figure 10.

[latex]x[/latex]	[latex]x^2[/latex]	[latex]-x^2[/latex]	Figure 10. Reflection across [latex]x[/latex]=axis. [latex](x, y)[/latex] transforms to [latex](x, -y)[/latex].
-3	9	-9
-2	4	-4
-1	1	-1
0	0	0
1	1	-1
2	4	-4
3	9	-9
Table 7. Reflecting the graph of [latex]f(x)=x^2[/latex] across the [latex]x[/latex]-axis transforms [latex]f(x)=x^2[/latex] into [latex]f(x)=-x^2[/latex]

Vertex Form of a Quadratic Function

All of the transformations we have applied to the function [latex]f(x)=x^2[/latex] can be combined. The result is the function [latex]f(x)=a(x-h)^2+k[/latex], where [latex]a[/latex] represents vertical stretching ([latex]a>1[/latex]) and compression ([latex]0<a<1[/latex]) as well as reflection across the [latex]x[/latex]-axis ([latex]a[/latex] is negative); [latex]h[/latex] represents a horizontal shift right when [latex]h>0[/latex] and left when [latex]h<0[/latex]; and [latex]k[/latex] represents a vertical shift up when [latex]k>0[/latex] and down when [latex]k<0[/latex]. In addition, the vertex will move from (0, 0) to [latex](h, k)[/latex].

Move the red dots in figure 11 to change the values of [latex]a, h[/latex], and [latex]k[/latex], and watch the transformation unfold.

Figure 11. Interactive transformation from [latex]f(x)=x^2[/latex] to [latex]f(x)=a(x-h)^2+k[/latex]

Determining the Equation of a Quadratic Function

The vertex form of a quadratic function is [latex]f(x)=a(x-h)^2+k[/latex], where [latex]a, h, k[/latex] are real numbers and [latex]a\ne 0[/latex]. So, if we know the values of [latex]a, h, k[/latex], we can write the corresponding function. For example, if we are told that  [latex]h=2,\;k=5,\;a=3[/latex]. Then the function is [latex]f(x)=3(x-2)^2+5[/latex]. Figure 12 shows the graph.

What if we were told instead that the vertex is (2, 5) and the graph passes through the point (1, 8)? Can we find the equation of the function knowing this information?

First off, we know that by definition the vertex gives us the values of [latex]h=2[/latex] and [latex]k=5[/latex]. This means that the function is [latex]f(x)=a(x-2)^2+5[/latex]. Now all we need is the value of [latex]a[/latex].Since the point (1, 8) lies on the graph, it must be a solution of the equation that represents the function, [latex]y=a(x-2)^2+5[/latex]. Substituting [latex]x=1,\;y=8[/latex] leaves [latex]a[/latex] as the only unknown so we can solve for [latex]a[/latex]:

[latex]\begin{aligned}y&=a(x-2)^2+5\\8&=a(1-2)^2+5\\8&=a+5\\3&=a\end{aligned}[/latex]