What is the vertex of $y = - 2 x^{2} + 2 x + 5$ $y=-2x^2 + 2x + 5$ ?

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What is the vertex of $y = - 2 x^{2} + 2 x + 5$ $y=-2x^2 + 2x + 5$ ?

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Answer 1 · 2017-08-10T12:50:57

$(\frac{1}{2}, \frac{11}{2})$ $(1/2,11/2)$

Explanation:

$given the equation of a parabola in standard form$ $"given the equation of a parabola in standard form"$

$that is y = a x^{2} + b x + c$ $"that is "y=ax^2+bx+c$

$then x_{vertex} = - \frac{b}{2 a}$ $"then "x_(color(red)"vertex")=-b/(2a)$

$y = - 2 x^{2} + 2 x + 5 is in standard form$ $y=-2x^2+2x+5" is in standard form"$

$with a = - 2, b = + 2, c = 5$ $"with "a=-2,b=+2,c=5$

$\Rightarrow x_{vertex} = - \frac{2}{- 4} = \frac{1}{2}$ $rArrx_(color(red)"vertex")=-2/(-4)=1/2$

$substitute this value into the equation for the corresponding$ $"substitute this value into the equation for the corresponding"$
$y-coordinate$ $"y-coordinate"$

< $\Rightarrow y_{vertex} = - 2 {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) + 5 = \frac{11}{2}$ $rArry_(color(red)"vertex")=-2(1/2)^2+2(1/2)+5=11/2$

$\Rightarrow vertex = (\frac{1}{2}, \frac{11}{2})$ $rArrcolor(magenta)"vertex "=(1/2,11/2)$

Answer 2 · 2017-08-10T12:54:58

Vertex is at $(\frac{1}{2}, \frac{11}{2})$ $(1/2, 11/2)$ .

Explanation:

The axis of symmetry is also the x value of the vertex. So we can use the formula $x = \frac{- b}{2 a}$ $x=(-b)/(2a)$ to find the axis of symmetry.

$x = \frac{- (2)}{2 (- 2)}$ $x=(-(2))/(2(-2))$

$x = \frac{1}{2}$ $x=1/2$

Substitute $x = \frac{1}{2}$ $x=1/2$ back into the original equation for the y value.

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

$y = \frac{11}{2}$ $y = 11/2$

Therefore, the vertex is at $(\frac{1}{2}, \frac{11}{2})$ $(1/2, 11/2)$ .

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What is the vertex of $y = - 2 x^{2} + 2 x + 5$ $y=-2x^2 + 2x + 5$ ?

2 Answers

Explanation:

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