How do you find the x intercepts for $f (x) = - x^{2} + 6 x + 8$ $f(x) = -x^2 + 6x + 8$ ?

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Answer 1 · 2015-06-03T20:59:19

To find x-intercepts, make y = 0. Solve f(x) = 0

-x^2 + 6x + 8 = 0
$D = d^{2} = 36 + 32 = 68 = 4 (17) \to d = \pm 2 \sqrt{17} .$ $D = d^2 = 36 + 32 = 68 = 4(17) -> d = +- 2sqrt17.$

$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = 2 \pm \sqrt{17}$ $x = -b/(2a) +- d/(2a) = 2 +- sqrt17$

How do you find the x intercepts for f(x) = -x^2 + 6x + 8f(x)=−x2+6x+8f(x) = -x^2 + 6x + 8?