Full solution

Q. The

xy

-plane shows the graph of the quadratic equation

y=-(\frac{1}{4})(x-3)^2+2

. Which of the following represents all solutions

(x,y)

to the system of equations created by this quadratic equation and the linear equation

y=\frac{1}{2}(x+3)-7

Given Equations: We are given two equations: $\newline$ $1$ . The quadratic equation $y = -\frac{1}{4}(x - 3)^2 + 2$ $\newline$ $2$ . The linear equation $y = \frac{1}{2}(x + 3) - 7$ $\newline$ To find the solutions to the system of equations, we need to set these two equations equal to each other and solve for $x$ .
Set Equations Equal: Set the quadratic equation equal to the linear equation: $\newline$ $-\frac{1}{4}(x - 3)^2 + 2 = \frac{1}{2}(x + 3) - 7$
Clear Fractions: Now we need to simplify and solve for $x$ . First, let's get rid of the fractions by multiplying every term by $4$ to clear the denominators: $\newline$ $-4 \times \left(\frac{1}{4}\right)(x - 3)^2 + 4 \times 2 = 4 \times \frac{1}{2}(x + 3) - 4 \times 7$
Simplify Equation: Simplify the equation: $- (x - 3)^2 + 8 = 2(x + 3) - 28$
Expand and Distribute: Next, we expand the left side of the equation and distribute the right side: $\newline$ - (x^\(2 - $6$ x + $9$ ) + $8$ = $2$ x + $6$ - $28$
Combine Like Terms: Distribute the negative sign on the left side and simplify the right side: $\newline$ $-x^2 + 6x - 9 + 8 = 2x - 22$
Move Terms and Solve: Combine like terms on the left side: $-x^2 + 6x - 1 = 2x - 22$
Quadratic Equation Solution: Now, we want to move all terms to one side to set the equation to zero and solve for $x$ : $-x^2 + 6x - 2x - 1 + 22 = 0$
Quadratic Equation Solution: Now, we want to move all terms to one side to set the equation to zero and solve for $x$ : $-x^2 + 6x - 2x - 1 + 22 = 0$ Combine like terms: $-x^2 + 4x + 21 = 0$
Quadratic Equation Solution: Now, we want to move all terms to one side to set the equation to zero and solve for $x$ : $-x^2 + 6x - 2x - 1 + 22 = 0$ Combine like terms: $-x^2 + 4x + 21 = 0$ This is a quadratic equation, and we can solve for $x$ by factoring, completing the square, or using the quadratic formula. Let's try to factor first: $(x - 3)(-x - 7) = 0$