Question

What is the vertex of this quadratic function? f(x) = 10x2 + 5x − 7 A. (0.25, -5.250) B. (0.25, -7.625) C. (-0.25, -5.250) D. (-0.25, -2.525) E. (-0.25, -7.625)

Answer 1

E

Given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then

The x-coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

y = 10x² + 5x - 7 is in standard form

with a = 10, b = 5 and c = - 7, hence

$x_{vertex}$ = - $\frac{5}{20}$ = - $\frac{1}{4}$

Substitute this value into the equation for y-coordinate

y = 10 (- $\frac{1}{4}$)² + 5 (- $\frac{1}{4}$) - 7

  = $\frac{5}{8}$ - $\frac{5}{4}$ - 7

  = $\frac{5}{8}$ - $\frac{10}{8}$ - $\frac{56}{8}$ = - $\frac{61}{8}$

vertex = (- 0.25, - 7.625 ) → E