How do you find the intercept and vertex of $y = - x^{2} + 9$ $y= -x^2 + 9$ ?

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Answer 1 · 2016-07-14T06:48:17

Vertex is $(0, 9)$ $(0, 9)$

Its y-intercept is $= 9$ $=9$

Given -

$y = - x^{2} + 9$ $y=-x^2+9$

It has no $x$ $x$ term. So we shall have it as -

$y = - x^{2} + 0 x + 9$ $y=-x^2+0x+9$

Then it is in the form $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

$x$ $x$ cor-ordinate of the vertex is -

$x = \frac{- b}{2 a} = \frac{0}{2 \times (- 1)} = 0$ $x=(-b)/(2a)=0/(2xx(-1))=0$

At $x = 0; y = (0^{2}) + 9 = 9$ $x=0; y=(0^2)+9=9$

Vertex is $(0, 9)$ $(0, 9)$

Its y-intercept is $= 9$ $=9$

Look at the graph

How do you find the intercept and vertex of y=−x2+9y= -x^2 + 9?