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Q. 39

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Found in: Page 644

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Find the vertex, focus, and directrix of each parabola. Graph the equation by hand. Verify your graph using a graphing utility.${y}^{2}=-16x$

The vertex is $\left(0,0\right)$ ,the focus is $\left(4,0\right)$ and directrix is $x=a,x=4$ .

The graph of the equation shown below.

See the step by step solution

## Step 1. Given information .

Consider the given equation ${y}^{2}=-16x$ .

## Step 2. Analyze the equation .

The standard form of parabola is ${y}^{2}=-4ax$

Compare the equation with standard form

.localid="1646816750877" ${y}^{2}=-4ax\phantom{\rule{0ex}{0ex}}{y}^{2}=-16x\phantom{\rule{0ex}{0ex}}-4a=-16\phantom{\rule{0ex}{0ex}}a=4$

Further simplify .

Since $a=4$ the vertex is $\left(0,0\right)$ ,the focus is $\left(a,0\right)=\left(4,0\right)$ and the directrix is localid="1646988458852" $x=4$

## Step 3. Plot the graph .

The graph of the equation ${y}^{2}=-16x$ using graphic utility is shown below .

Here V is the vertex , F is the focus and line $x=4$ represents the directrix of the parabola .