How do you find the vertex and intercepts for $f (x) = - 4 x^{2} + 4 x + 4$ $f(x)= -4x^2 + 4x + 4$ ?

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How do you find the vertex and intercepts for $f (x) = - 4 x^{2} + 4 x + 4$ $f(x)= -4x^2 + 4x + 4$ ?

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Answer 1 · 2018-06-08T05:43:00

Vertex $(- \frac{1}{2}, 5)$ $(-1/2, 5)$
Y - intercept $(0, 4)$ $(0, 4)$
X-intercept $(\frac{\sqrt{5} + 1}{2}, 0); (\frac{- \sqrt{5} + 1}{2}, 0)$ $( (sqrt5+1)/2,0); ((-sqrt5+1)/2,0)$

Explanation:

Given -

$f (x) = - 4 x^{2} + 4 x + 4$ $f(x)=-4x^2+4x+4$

$y = - 4 x^{2} + 4 x + 4$ $y=-4x^2+4x+4$

Vertex

$x = \frac{- b}{2 a} = \frac{- (4)}{2 \times - 4} = \frac{- 4}{- 8} = \frac{1}{2}$ $x=(-b)/(2a)=(-(4))/(2xx-4)=(-4)/-8=1/2$

At $x = - \frac{1}{2}; y = - 4 {(\frac{1}{2})}^{2} + 4 (\frac{1}{2}) + 4 = - 1 + 2 + 4 = 5$ $x=-1/2; y=-4(1/2)^2+4(1/2)+4=-1+2+4=5$

$(\frac{1}{2}, 5)$ $(1/2, 5)$

At $x = 0; y = - 4 {(0)}^{2} + 4 (0) + 4 = 4$ $x=0;y=-4(0)^2+4(0)+4=4$

Y - intercept $(0, 4)$ $(0, 4)$

X-intercept

At $y = 0; - 4 x^{2} + 4 x + 4 = 0$ $y=0; -4x^2+4x+4=0$

$- 4 x^{2} + 4 x + 4 = 0$ $-4x^2+4x+4=0$

Dividing both sides by $- 4$ $-4$

$x^{2} - x - 1 = 0$ $x^2-x-1=0$

$x^{2} - x = 1$ $x^2-x=1$

$x^{2} - x + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}$ $x^2-x+1/4=1+1/4=5/4$

${(x - \frac{1}{2})}^{2} = \frac{5}{4})$ $(x-1/2)^2=5/4)$

$x - \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$ $x-1/2=+-sqrt(5/4)=+-sqrt5/2$

$x = \frac{\sqrt{5}}{2} + \frac{1}{2} = \frac{\sqrt{5} + 1}{2}$ $x=sqrt5/2+1/2=(sqrt5+1)/2$

$x = - \frac{\sqrt{5}}{2} + \frac{1}{2} = \frac{- \sqrt{5} + 1}{2}$ $x=-sqrt5/2+1/2=(-sqrt5+1)/2$

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How do you find the vertex and intercepts for $f (x) = - 4 x^{2} + 4 x + 4$ $f(x)= -4x^2 + 4x + 4$ ?

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Explanation:

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How do you find the vertex and intercepts for f(x)=−4x2+4x+4f(x)= -4x^2 + 4x + 4?

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How do you find the vertex and intercepts for $f (x) = - 4 x^{2} + 4 x + 4$ $f(x)= -4x^2 + 4x + 4$ ?