Question

use graphing to find the solutions to the system of equations.

1. \

$$\begin{cases} y = x^2 + 7x + 7 \\\\ \\ \\ \\ y = x + 2 \\end{cases}$$

a.
\includegraphicsscale=0.5{graph_a}
(-3, 3)
(-1, -1)
b.
\includegraphicsscale=0.5{graph_b}
(-5, -3)
(-1, 1)
c.
\includegraphicsscale=0.5{graph_c}
(-4, -3)
(-2, -1)
d.
\includegraphicsscale=0.5{graph_d}
(-4, 3)
(-2, 1)

Explanation:

Step1: Analyze the first equation

The equation \(y = x^{2}+7x + 7\) is a quadratic function, so its graph is a parabola opening upwards (since the coefficient of \(x^{2}\) is positive, \(a = 1>0\)). The linear equation \(y=x + 2\) has a slope of \(1\) (positive) and a \(y\)-intercept of \(2\).

Step2: Check the intersection points

We can also solve the system algebraically to find the intersection points and then match with the graphs. Set \(x^{2}+7x + 7=x + 2\).
Subtract \(x\) and \(2\) from both sides: \(x^{2}+6x + 5 = 0\).
Factor the quadratic: \((x + 1)(x+5)=0\).
So \(x=-1\) or \(x=-5\).
When \(x=-1\), \(y=-1 + 2=1\). When \(x=-5\), \(y=-5 + 2=-3\). So the solution points are \((-5,-3)\) and \((-1,1)\).

Now check the graphs:

• Graph a: The parabola opens downward (incorrect, our parabola opens upward), so eliminate a.
• Graph b: The intersection points are \((-5,-3)\) and \((-1,1)\), which matches our solution.
• Graph c: The intersection points given are \((-4,-3)\) and \((-2,-1)\), which do not match.
• Graph d: The intersection points given are \((-4,3)\) and \((-2,1)\), which do not match.

Answer:

b. The graph with intersection points \((-5, -3)\) and \((-1, 1)\)