# How do you find the vertex and intercepts for  f(x) = -7x^2 + 3x + 1?

Jul 27, 2018

Vertex $\left(\frac{3}{14} , \frac{37}{28}\right)$
y-intercept $\left(0 , 1\right)$

x-intercept ((sqrt37+3)/14,0); ((-sqrt37+3)/14, 0)

#### Explanation:

Give -

$f \left(x\right) = - 7 {x}^{2} + 3 x + 1$

$y = - 7 {x}^{2} + 3 x + 1$

Vertex

$x = \frac{- b}{2 a} = \frac{- 3}{2 \times - 7} = \frac{- 3}{- 14} = \frac{3}{14}$

At x=3/14; y=-7(3/14)^2+3(3/14)+1

$y = - 7 \left(\frac{9}{196}\right) + \frac{9}{14} + 1 = \frac{- 63 + 126 + 196}{196} = \frac{259}{196} = \frac{37}{28}$

$\left(\frac{3}{14} , \frac{37}{28}\right)$

y-intercept

At x=0; y=-7(0)+3(0)+1=1

y-intercept $\left(0 , 1\right)$

x-intercept

Aty=0; -7x^2+3x+1=0

$- 7 {x}^{2} + 3 x = - 1$

${x}^{2} - \frac{3}{7} x + \frac{9}{196} = - \frac{1}{- 7} + \frac{9}{196} = \frac{28 + 9}{196} = \frac{37}{196}$

${\left(x - \frac{3}{14}\right)}^{2} = \pm \sqrt{\frac{37}{196}} = \pm \frac{\sqrt{37}}{14}$

$x = \frac{\sqrt{37}}{14} + \frac{3}{14} = \frac{\sqrt{37} + 3}{14}$
$x = - \frac{\sqrt{37}}{14} + \frac{3}{14} = \frac{- \sqrt{37} + 3}{14}$

x-intercept ((sqrt37+3)/14,0); ((-sqrt37+3)/14, 0)