Question

Rewrite the quadratic function by completing the square. State the x intercepts


f(x) = 4x2 + 8x + 7

Answer 1

do you have to solve for x?

Answer 2

yes

Answer 3

Help

Answer 4

do you know what a "perfect square trinomial" is?

Answer 5

Yes, but I am having difficulties with this one

Answer 6

well... ok... then let's start off by grouping the "x's" so \(\bf f(x) = 4x^2 + 8x + 7\implies f(x) = (4x^2 + 8x) + 7\\ \quad \\ f(x) = 4(x^2 + 2x) + 7 \quad \implies
f(x) = 4(x^2 + 2x+\quad \square^2\quad ) + 7\)


so... what do you think is our missing number there to get a perfect square trinomial?

Answer 7

1

Answer 8

1.... so we'll add 1 then
keep in mind that, all we're doing is borrowing from our good fellow Zero, 0, so if we add that much, we also have to subtract it, notice the 4 outside as common factor, so \(\bf f(x) = 4(x^2 + 2x+1^2\quad ) + 7\\\quad \textit{we added}\quad 4(1^2)\quad \textit{we have to subtract }\quad 4(1^2)\\ \quad \\
f(x) = 4(x^2 + 2x+1^2)+7 - 4(1^2)\implies f(x) = 4(x+1)^2+7-4\)

Answer 9

okay, so what would the x intercept be?

Answer 10

well, for the x-intercept, just set y = f(x) = 0 and solve for "x" \(\bf 0 = 4(x+1)^2+7-4\)

Answer 11

\(\bf 0 = 4(x+1)^2+3\implies -3 = 4(x+1)^2\implies \cfrac{-3}{4} = (x+1)^2\\ \quad \\
\pm\sqrt{\cfrac{-3}{4}} = x+1\implies \pm\sqrt{\cfrac{-3}{4}}-1 = x\)

so the root will give you 2 values, just like you'd have if using the quadratic formula

Answer 12

hmm well. the radical value is negative... it just means 2 complex values

Answer 13

I wrote the problem wrong it is -7 not +7 how does that change my answer

Answer 14

well... then \(\bf f(x) = 4(x^2 + 2x+1^2) - 7\\\quad \textit{we added}\quad 4(1^2)\quad \textit{we have to subtract }\quad 4(1^2)\\ \quad \\ f(x) = 4(x^2 + 2x+1^2)-7 - 4(1^2)\implies f(x) = 4(x+1)^2-7-4\)

Answer 15

does that change the x intercept and how would I graph it

Answer 16

that does change the x-intercept, yes,, but you'd do the same, set y = 0 and solve for "x"

how to graph it? well, easy, the idea behind simplifying it to use a perfect square trinomial is to make the "vertex form" -> http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

Answer 17

When I graph using my graphing calculator it puts this point at -11/4. Is this correct

Answer 18

well. let's see that a parabola "focus form" equation will look like \(\bf x-h)^2=4p(y-k)\) where the vertex is at (h, k) so, if you change yours to that, you'll notice the vertex


ohh I see what you mean... the multiplier 4.. yes... . the vertex form will work better like \(\bf x-h)^2=4p(y-k)\) my bad

Answer 19

So that -11/4 point is correct

Answer 20

?

Answer 21

\(\bf y= 4(x+1)^2-11\implies (y-(\color{red}{-11}))=4(x-(\color{red}{-1}))^2\)

Answer 22

using the form of \(\bf \large (x-h)^2=4p(y-k)\)

Answer 23

okay?

Answer 24

so yes, the equation is fine