What is the vertex of $y = - {(x + 1)}^{2} + 17$ $y= -(x+1)^2 +17$ ?

Question

1759 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-16T05:00:43

vertex $= (- 1, 17)$ $=(-1,17)$

The general equation of a quadratic equation in vertex form is:

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

where:
$a =$ $a=$ vertical stretch/compression
$h =$ $h=$ x-coordinate of vertex
$k =$ $k=$ y-coordinate of vertex

Looking back at the equation, $y = - {(x + 1)}^{2} + 17$ $y=-(x+1)^2+17$ , we can see that:

Keep in mind that $h$ $h$ is negative and not positive even though it appears to be in the equation.

$:.$ , the vertex is $(-1,17)$ .

Answer 2 · 2015-12-16T05:01:32

Vertex (-1, 17)

y is expressed in vertex form.
Vertex (-1, 17)

What is the vertex of y= -(x+1)^2 +17 y=−(x+1)2+17y= -(x+1)^2 +17 ?