What is the vertex of $y = - 2 {(x - 3)}^{2} - 1$ $y= -2(x-3)^2-1$ ?

Question

3592 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-15T15:37:33

Vertex $\to (x, y) = (3, - 1)$ $->(x,y)=(3,-1)$

When the quadratic equation is in this form you can almost read the coordinates of the vertex strait off. It just needs a little tweaking.

Suppose we wrote it as $y = a {(x + d)}^{2} + f$ $y=a(x+d)^2+f$

Then the vertex $\to (x, y) = (- d, f)$ $->(x,y)=(-d,f)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the format of the above example we have:

Vertex $\to (x, y) = (3, - 1)$ $->(x,y)=(3,-1)$

Tony B

What is the vertex of y=−2(x−3)2−1y= -2(x-3)^2-1?