Question

name:
date:
per:
unit 4: solving quadratic equations
homework 12: solving nonlinear systems graphically
this is a 2 - page document!
solve each system by graphing. be sure to identify the solution(s).

1. (\begin{cases}y = 4x + 5\\y = x^{2}+8x + 9end{cases})
2. (\begin{cases}y=-x^{2}-4x + 2\\y = x^{2}+8x + 12end{cases})
3. (\begin{cases}y = x^{2}+12x + 26\\y=-x^{2}-4x - 6end{cases})
4. (\begin{cases}y=-x^{2}+10x - 28\\x - y = 7end{cases})
5. (\begin{cases}y = x^{2}-2x - 7\\y=-3x - 5end{cases})
6. (\begin{cases}y = x^{2}-6x + 10\\y=-2x^{2}+4xend{cases})

Explanation:

Problem 1:

Step1: Set equations equal

To find the solution, set \(4x + 5=x^{2}+8x + 9\).

Step2: Rearrange into quadratic

Rearrange to \(x^{2}+4x + 4 = 0\).

Step3: Factor the quadratic

Factor: \((x + 2)^{2}=0\), so \(x=-2\).

Step4: Find y - value

Substitute \(x = - 2\) into \(y = 4x+5\): \(y=4(-2)+5=-3\).

Step1: Set equations equal

Set \(-x^{2}-4x + 2=x^{2}+8x + 12\).

Step2: Rearrange into quadratic

Rearrange to \(2x^{2}+12x + 10 = 0\), divide by 2: \(x^{2}+6x + 5 = 0\).

Step3: Factor the quadratic

Factor: \((x + 1)(x + 5)=0\), so \(x=-1\) or \(x=-5\).

Step4: Find y - values

For \(x=-1\): \(y=-(-1)^{2}-4(-1)+2=-1 + 4+2 = 5\). For \(x=-5\): \(y=-(-5)^{2}-4(-5)+2=-25 + 20+2=-3\).

Step1: Set equations equal

Set \(x^{2}+12x + 26=-x^{2}-4x - 6\).

Step2: Rearrange into quadratic

Rearrange to \(2x^{2}+16x + 32 = 0\), divide by 2: \(x^{2}+8x + 16 = 0\).

Step3: Factor the quadratic

Factor: \((x + 4)^{2}=0\), so \(x=-4\).

Step4: Find y - value

Substitute \(x=-4\) into \(y=-x^{2}-4x - 6\): \(y=-(-4)^{2}-4(-4)-6=-16 + 16-6=-6\).

Answer:

The solution is \((-2,-3)\)

Problem 2: