Given $f(x)= \frac{1}{4}(x+4)^2-2$ Find vertex, $ y$ intercept etc.

Question

Given $f(x)= \frac{1}{4}(x+4)^2-2$

Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry

What I have so far:

Vertex: $(-4,-2)$

$y$-intercept: $(0,2)$

$x$-intercept: $2$

Axis of symmetry $x=-4$

If you could please tell me if these are right or not that would be great. If they are not right please correct them and tell me why they are wrong.

Answer 1

Everything is right except for the x intercept(s). Try moving the -2 to the Left Side, and then simplyfying to get x in terms of square roots.

Answer 2

Because the function is in parabola vertex-form the vertex is at $\left( -4,-2\right)$

The axis of symmetry is about the $x-$value of the vertex, because the function is a vertical parabola. $$ \therefore x_{axis}=-4 $$

for the $y-$intercept you need $f\left( 0\right)$ $$ f\left( 0\right) = \frac{1}{4}\left( 0+4\right) ^2-2 $$ $$ f\left( 0\right) =\frac{4^2}{4}-2 $$ $$ f\left( 0\right) = 4-2 $$ $$ f\left( 0\right) = 2 $$

For $x-$intercepts the line $y=0$ intersects the function $y=\frac{1}{4}\left( x+4\right) ^2-2$

$$\therefore \quad 0=\frac{1}{4}\left( x+4\right) ^2-2$$ $$ 2= \frac{1}{4}\left( x+4\right) ^2$$ $$ 8= \left( x+4\right) ^2$$ $$ x+4=\pm\sqrt8 $$ $$ x=-4\pm\sqrt8 $$

Answer 3

Consider $y=a(x+4)^2-2$. Let $Y=y+2$ and $X=x+4$. Then $Y=aX^2$.

This means that the curve $y=a(x+4)^2-2$ is a translation of the curve $Y=aX^2$: the origin $(0,0)$ in $XY$-space goes to the point $(-4,-2)$ in $xy$-space.

Thus, the vertex of $y=a(x+4)^2-2$ is at $(-4,-2)$, the symmetry axis is $x=-4$, etc.