Question

Can someone help with Quadratic Functions?

Find the solutions to the system
y=x^2+8x+2
y=7x+4

I know first to set them and then from there but my difficulty is with expanding them. Once I get to there I get stuck. Please Help.

Answer 1

@CookieCake

Answer 2

Okay, for the first I got (-4,-14)

How?

Y=X^2+8X+2
X^2+8X+2=0
A=1 B=8 C=2

-b/2a (just plug the A and B into the formula.)

-b/2a= -8/2(1) = -8/2= -4/1=-4 Axis of symmetry and X axis of vertex.
X=-4


C-b^2/4a

(2)-8^2/4(1) = 2-64/4= 2-16= -14

Y=-14

(-4,-14)

Answer 3

The correct answer is (-2,-10) & (1,11) the thing is I dont know how and I know this is right for a fact.

Answer 4

I'm sorry if my work did not help. I recommend a graphing calculator, or online math solver to check your work.

Best of wishes!

Again,sorry.

Answer 5

Its fine & I have tried a solver and it cant help me I really dont get it & generally i'm alright with this kind of stuff. Thank you for trying, thats alot and I am really gracious for the time you did put into it.

Answer 6

\[x^2+8x+2 =7x+4 \] right?

Answer 7

set equal to zero and get
\[x^2+x-2=0\] then factor as
\[(x+2)(x-1)=0\] and now you can solve for \(x\) easily

Answer 8

have no idea how the other answer came about, the answer is \(\{-2,1\}\)

Answer 9

Yes your right and you set them equal to zero. But wouldnt that just be the x's? the -10 and the 11 would be the y's. SO your correct for the x's part.

Answer 10

So**

Answer 11

Yep -10 for 7*-2+4

Answer 12

yeah you can find the second coordinate by replacing \(x\) by \(-2\) or by replacing \(x\) by \(1\)
you get the same y because you are solving for where they are equal

Answer 13

Thats all they are wanting. Thank you so much though for the expanding part I was so very confused. I will fan and medal you.