How do you find the vertex and the intercepts for $f(x) = -(x+5)^2$ ?

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Answer 1 · 2017-01-17T17:06:49

The vertex form for the equation of a parabola such that $y = f(x)$ is:
$y = a(x - h)^2 + k" [1]"$

where $(h,k)$ is the vertex.

Write the given equation in the form of equation [1]:

$y = -(x - -5)^2" [2]"$

The vertex is the point $(-5, 0)$

To find the x intercept(s), set equation [2] equal to zero and solve for x:

$0 = -(x - -5)^2$

$x = -5$

Because the vertex is, also, the x intercept, there is only one x intercept at the point $(-5, 0)$ .

To find the y intercept set x, in equation [2], equal to zero and solve for y:

$y = -(-5)^2$

$y = -25$

The y intercept is the point $(0, -25)$

How do you find the vertex and the intercepts for f(x) = -(x+5)^2f(x) = -(x+5)^2?