# How do you find the vertex and intercepts for f(x)=-9x^2 + 7x + 6?

Apr 16, 2017

Vertex is at $\left(0.39 , 7.36\right)$, x-intercept is at $\left(1.29 , 0\right) \mathmr{and} \left(- 0.52 , 0\right)$ , y-intercept is at $\left(0 , 6\right)$

#### Explanation:

f(x) = -9x^2 +7x +6 ; a= -9 , b=7 , c=6 ,  comparing with standard equation $f \left(x\right) = a {x}^{2} + b x + c$.

Vertex (x-ordinate) $= - \frac{b}{2 a} = - \frac{7}{-} 18 = \frac{7}{18} \approx 0. 39 \left(2 \mathrm{dp}\right)$
Vertex (y-ordinate) $f \left(x\right) = - 9 \cdot {7}^{2} / {18}^{2} + 7 \cdot \frac{7}{18} + 6 \approx 7.36 \left(2 \mathrm{dp}\right)$

So Vertex is at $\left(0.39 , 7.36\right)$

y-intercept can be obtained by putting $x = 0$ in the equation , i.e $f \left(x\right) = - 9 \cdot 0 + 7 \cdot 0 + 6 = 6 \therefore$ y-intercept is at $\left(0 , 6\right)$

x-intercept can be obtained by putting $f \left(x\right) = 0$ in the equation , i.e
$0 = - 9 {x}^{2} + 7 x + 6$
$\therefore x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$
$\therefore x = - \frac{7}{-} 18 \pm \frac{\sqrt{49 - 4 \cdot - 9 \cdot 6}}{-} 18 = \frac{7}{18} \pm \frac{\sqrt{265}}{-} 18$
$\therefore x \approx - 0.52 \left(2 \mathrm{dp}\right) , x \approx 1.29 \left(2 \mathrm{dp}\right)$

x-intercept is at $\left(1.29 , 0\right) \mathmr{and} \left(- 0.52 , 0\right)$ graph{-9x^2+7x+6 [-20, 20, -10, 10]}[Ans]