What is the vertex of $y = x^{2} - 9 x + 14$ $y= x^2-9x+14$ ?

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What is the vertex of $y = x^{2} - 9 x + 14$ $y= x^2-9x+14$ ?

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Answer 1 · 2018-04-05T17:16:03

(4.5,-4.9)

Explanation:

$a x^{2} + b x + c$ $ax^2+bx+c$ is the general quadratic equation and $- \frac{b}{2 a}$ $-b/(2a)$ will give the X coordinate of the line of symmetry/the maximum or minimum point.

Substitute this value into the equation to find the $y$ $y$ value

$x^{2} - 9 x + 14$ $x^2-9x+14$ $\Rightarrow$ $=>$ . $\frac{- - 9}{2}$ $(--9)/2$ = $\frac{9}{2}$ $9/2$ =4.5

${(4.5)}^{2} - 9 \times 4.5 + 14$ $(4.5)^2-9xx4.5+14$

=-4.9

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What is the vertex of $y = x^{2} - 9 x + 14$ $y= x^2-9x+14$ ?

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What is the vertex of y= x^2-9x+14 y=x2−9x+14y= x^2-9x+14 ?

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What is the vertex of $y = x^{2} - 9 x + 14$ $y= x^2-9x+14$ ?