How do you find the vertex of $y=x^2-8x+18$ ?

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Answer 1 · 2015-12-18T06:44:22

The vertex is $(4,2)$ .

$y=x^2-8x+18$ is a quadratic equation in the form $ax^2+bx+c$ , where $a=1, b=-8, and c=18$ .

The vertex is the maximum or minimum point of a parabola.

To find the $x$ value of the vertex use the formula $x=(-b)/(2a)$ .

$x=(-(-8))/(2*1)$

$x=8/2$

$x=4$

To find the value of $y$ , substitute the value for $x$ into the equation.

$y=x^2-8x+18$

$y=4^2-(8*4)+18$

$y=16-32+18$

$y=2$

The vertex is $(4,2)$ .

graph{y=x^2-8x+18 [-9.944, 10.06, -0.07, 9.93]}

How do you find the vertex of y=x^2-8x+18y=x^2-8x+18?