How do you find the vertex and the intercepts for $n(x) = -8x² + 8x - 3$ ?

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Answer 1 · 2016-06-28T06:41:39

Vertex $(1/2, 3/4)$
Intercept $(0, -3)$

Given -

$y=-8x^2+8x-3$

Vertex -

$x=(-b)/(2a)=(-8)/(2xx(-8))=(-8)/(-16)=1/2$
At $x=1/2$
$y=-(1/2)^2+8(1/2)-3$
$y=-1/4+4-3=3/4$

$(1/2, 3/4)$

Intercept:

At $x=0$

$y=-8(0)+8(0)-3=-3$
$y=-3$

How do you find the vertex and the intercepts for n(x) = -8x² + 8x - 3n(x) = -8x² + 8x - 3?