What is the vertex form of $y = x^{2} - 14 x + 16$ $y = x^2 -14x + 16$ ?

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Answer 1 · 2018-03-14T18:15:45

$y = {(x - 7)}^{2} - 33$ $y=(x-7)^2-33$

First find the vertex using the formula
$x = \frac{- b}{2a}$ $x=(-b)/"2a"$

a=1
b=-14
c=16

$x = \frac{- (- 14)}{2(1)}$ $x=(-(-14))/"2(1)"$ This simplifies to $x = \frac{14}{2}$ $x=14/"2"$ which is 7.
so $x = 7$ $x=7$

So on now that we have x we can find y.

$y = x^{2} - 14 x + 16$ $y=x^2-14x+16$
$y = {(7)}^{2} - 14 (7) + 16$ $y=(7)^2-14(7)+16$
$y = - 33$ $y=-33$

$V e r t e x = (7, - 33)$ $Vertex = (7,-33)$ where h=7 and k=-33

We now finally enter this into vertex form which is,
$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

x and y in the "vertex form" are not associated with the values we found earlier.

$y = 1 {(x - 7)}^{2} + (- 33)$ $y=1(x-7)^2+(-33)$
$y = {(x - 7)}^{2} - 33$ $y=(x-7)^2-33$

What is the vertex form of y=x2−14x+16y = x^2 -14x + 16?