How do you find the vertex and intercepts for f(x)= -x^2 + 6x + 3?

Jan 8, 2017

The vertex is $= \left(3 , 6\right)$
The intercepts are $\left(0 , 3\right)$, $\left(3 + 2 \sqrt{3} , 0\right)$ and $\left(3 - 2 \sqrt{3} , 0\right)$

Explanation:

The vertex is the highest point or lowest point of the equation,

we complete the squares

We rewrite $f \left(x\right)$ as

$f \left(x\right) = - {x}^{2} + 6 x + 3$

$= - \left({x}^{2} - 6 x\right) + 3$

$= - \left({x}^{2} - 6 x + 9\right) + 3 + 9$

$= - {\left(x - 3\right)}^{2} + 12$

We compare this to $f \left(x\right) = a {\left(x - h\right)}^{2} + k$

and the vertex is $\left(h , k\right) = \left(3 , 6\right)$

The axis of symmetry is $x = 3$

To find the intercepts,

first,

Let $x = 0$, $\implies$, $\left(y = 3\right)$

second,

Let $y = 0$, then

$- {\left(x - 3\right)}^{2} + 12 = 0$

${\left(x - 3\right)}^{2} = 12$

$x - 3 = \pm \sqrt{12} = \pm 2 \sqrt{3}$

$x = 3 \pm 2 \sqrt{3}$

Therefore,

The points are $\left(3 + 2 \sqrt{3} , 0\right)$ and $\left(3 - 2 \sqrt{3} , 0\right)$

The points are $\left(6.464 , 0\right)$ and $\left(- 0.464 , 0\right)$

graph{(y-(-x^2+6x+3))((x-3)^2+(y-12)^2-0.01)(y-1000x+3000)=0 [-18.1, 17.93, -2.36, 15.67]}