# Thread: Horizontal stretch by a factor of 5

1. ## Horizontal stretch by a factor of 5

Will you please perform horizontal stretch by a factor of 5 of the parent function y=3x^2+5x

Thank you very much

2. Originally Posted by oceanmd
Will you please perform horizontal stretch by a factor of 5 of the parent function y=3x^2+5x

Thank you very much
A horizontal stretch means we effect the x-axis.

To stretch the x-axis by a factor of 5, we "paradoxically" alter x by a factor of 1/5:
$y' = 3 \left ( \frac{x}{5} \right ) ^2 + 5 \left ( \frac{x}{5} \right )$

$y' = \frac{3}{25}x^2 + x$

-Dan

3. Dan,

thank you very much, you confirmed what I did, the only difference is that when I graphed both parabolas, the vertex for foth functions is at (0,0). Why have you graphed them differently?

4. Originally Posted by oceanmd
Dan,

thank you very much, you confirmed what I did, the only difference is that when I graphed both parabolas, the vertex for foth functions is at (0,0). Why have you graphed them differently?
The vertex for $y=3x^2+5x$ can be found by completing the square:
$y=3x^2+5x$

$y = 3 \left ( x^2 + \frac{5}{3}x \right )$

$y = 3 \left ( x^2 + \frac{5}{3}x + \frac{25}{36} - \frac{25}{36} \right )$

$y = 3 \left ( x^2 + \frac{5}{3}x + \frac{25}{36} \right ) - 3 \cdot \frac{25}{36}$

$y = 3 \left ( x + \frac{5}{6} \right )^2 - \frac{25}{12}$

So the vertex is at $\left ( -\frac{5}{6}, -\frac{25}{12} \right )$ as my graph shows.

-Dan

5. Dan,

Thanks, it is just hard to keep everything in mind. Axis od symmetry formula -b/2a will give x=-5/6

Thank you so much