Question

Please Help!

Given the quadratic equation: x2 - 8x + 11 = 0

Write the quadratic function in vertex form.

Answer 1

To get a quadratic in vertex form you need to "complete the square". This entails rearranging it into the form of a(x-b)^2+c

So we put the x^2 and -8x in a bracket \[(x^2-8x)+11=0\]

Now we are going to move the x's square outside the bracket, but first we need to half the -8 and take it's x off so if we were to expand the bracket later it would form a -8x again.

\[(x-4)^{2}+11=0\]

If we were to expand that bracket now sure we would get our -8x, but we would also get a random 16 popping out that wasn't there before, so to counter act this we tag -16 onto the end\[(x-4)^{2}+11-16=0\]Which simplifies to\[(x-4)^{2}-5=0\] giving us our answer :)

Answer 2

y = x² + 8x + 11
line of symmetry is x = -8/2 = -4
y = 4² + 8(-4) + 11 = 16 – 32 + 11 = -5
so vertex is (-4, -5)

since x axis is 5 up from vertex, and for quads, up is over squared,
zeros at -4 ± √5

x² + 8x + 11 = 0
x² + 8x + ... = -11 ∙ ∙ ∙ ∙ ∙ ∙ add (-8/2)²
x² + 8x + 16 = -11 + 16
(x + 4)² = 5
x + 4 = ±√5
x = -4 ± √5