Question

(a) use transformations to get the graph of $y = -\frac{1}{2}x^2$.
(b) use transformations to get the graph of $y = (2x)^2$.

Explanation:

Step1: Identify parent function

The parent function is $y=x^2$, graphed above.

Step2: Transform for $y=-\frac{1}{2}x^2$

1. Vertical compression: Multiply $y=x^2$ by $\frac{1}{2}$ to get $y=\frac{1}{2}x^2$. This shrinks the graph vertically by a factor of $\frac{1}{2}$ (each $y$-value is halved).
2. Reflection over x-axis: Multiply $y=\frac{1}{2}x^2$ by $-1$ to get $y=-\frac{1}{2}x^2$. This flips the graph over the $x$-axis, so it opens downward.

Key points for $y=-\frac{1}{2}x^2$:

• When $x=0$, $y=0$
• When $x=\pm1$, $y=-\frac{1}{2}(1)^2=-\frac{1}{2}$
• When $x=\pm2$, $y=-\frac{1}{2}(4)=-2$
• When $x=\pm4$, $y=-\frac{1}{2}(16)=-8$

Step3: Transform for $y=(2x)^2$

Horizontal compression: Replace $x$ with $2x$ in $y=x^2$ to get $y=(2x)^2=4x^2$. This shrinks the graph horizontally by a factor of $\frac{1}{2}$ (the graph reaches the same $y$-values at half the $x$-input of the parent function).

Key points for $y=(2x)^2$:

• When $x=0$, $y=0$
• When $x=\pm0.5$, $y=(2(\pm0.5))^2=(\pm1)^2=1$
• When $x=\pm1$, $y=(2(\pm1))^2=(\pm2)^2=4$
• When $x=\pm2$, $y=(2(\pm2))^2=(\pm4)^2=16$

Answer:

(a) The graph of $y=-\frac{1}{2}x^2$ is a downward-opening parabola, vertically compressed by a factor of $\frac{1}{2}$ relative to $y=x^2$, passing through points $(0,0)$, $(\pm1, -\frac{1}{2})$, $(\pm2, -2)$.
(b) The graph of $y=(2x)^2$ is an upward-opening parabola, horizontally compressed by a factor of $\frac{1}{2}$ relative to $y=x^2$, passing through points $(0,0)$, $(\pm0.5, 1)$, $(\pm1, 4)$.