How do you graph y=-2(3+x)^2 +4?

Aug 6, 2015

Calculate the vertex and the $x$- and $y$-intercepts and then sketch the graph.

Explanation:

$y = - 2 {\left(3 + x\right)}^{2} + 4$

Step 1. Convert your equation to vertex form.

The vertex form is $y = a {\left(x - h\right)}^{2} + k$.

Rearrange your equation to $y = - 2 {\left(x + 3\right)}^{2} + 4$.

Then $a = - 2$, $h = - 3$, and $k = 4$.

Step 2. Find the vertex.

The vertex is at ($h , k$) or ($- 3 , 4$).

Step 3. Find the $y$-intercept.

Set $x = 0$ and solve for $y$.

$y = - 2 {\left(3 + x\right)}^{2} + 4 = - 2 {\left(3\right)}^{2} + 4 = - 2 \left(9\right) + 4 = - 18 + 4 = - 14$

The $y$-intercept is at ($0 , - 14$).

Step 4. Find the $x$-intercept(s).

Set $y = 0$ and solve for $x$.

$0 = - 2 {\left(3 + x\right)}^{2} + 4$

${\left(3 + x\right)}^{2} - 2 = 0$

$\left(\left(3 + x\right) + \sqrt{2}\right) \left(\left(3 + x\right) - \sqrt{2}\right) = 0$

$3 + x + \sqrt{2} = 0$ and $3 + x - \sqrt{2} = 0$

x=-3-sqrt2≈ -4.4 and x=-3+sqrt2≈-1.6

Step 5. Draw your axes and plot the four points.

Step 6. Draw a smooth parabola that passes through the four points.