What is the vertex of $y = x^{2} - 4 x - 3$ $y= x^2 - 4x - 3$ ?

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What is the vertex of $y = x^{2} - 4 x - 3$ $y= x^2 - 4x - 3$ ?

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Answer 1 · 2018-04-08T08:58:48

$(2, - 7)$ $(2,-7)$

Explanation:

$\frac{- b}{2 a}$ $(-b)/(2a)$ is the $x$ $x$ value for the maximum/minimum (vertex) of a quadratic graph. Work out what this value is and put it into the equation to find the $y$ $y$ value.

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

$\Rightarrow$ $=>$ $y = 2^{2} - 4 \times 2 - 3$ $y=2^2-4xx2-3$

$\Rightarrow$ $=>$ $y = 4 - 8 - 3$ $y=4-8-3$

$y = - 7$ $y=-7$

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What is the vertex of $y = x^{2} - 4 x - 3$ $y= x^2 - 4x - 3$ ?

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What is the vertex of y= x^2 - 4x - 3y=x2−4x−3 y= x^2 - 4x - 3?

1 Answer

Explanation:

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What is the vertex of $y = x^{2} - 4 x - 3$ $y= x^2 - 4x - 3$ ?