Question

solve the system of equations by graphing. first graph the equations, and then identify the solutions.
\\( y = -5x + 4 \\)
\\( y = -(x + 5)^2 + 1 \\)
to graph a linear equation, click to plot points on the graph. to graph a quadratic equation, click to plot the vertex first, then plot another point on the graph.
graph with x-axis from -10 to 10 and y-axis from -10 to 10, with linear, quadratic, and edit buttons below
what are the solutions to the system of equations? select all that apply.
(1, -1) (-4, 0) (0, 4)
(-5, 1) there are no solutions.

Explanation:

Step1: Check (1, -1)

Substitute \(x = 1\), \(y=-1\) into \(y=-5x + 4\): \(-1=-5(1)+4=-1\), true.
Substitute into \(y=-(x + 5)^2+1\): \(-1=-(1 + 5)^2+1=-36 + 1=-35\), false.

Step2: Check (-4, 0)

Substitute \(x=-4\), \(y = 0\) into \(y=-5x + 4\): \(0=-5(-4)+4=24\), false.

Step3: Check (0, 4)

Substitute \(x = 0\), \(y = 4\) into \(y=-5x + 4\): \(4=-5(0)+4=4\), true.
Substitute into \(y=-(x + 5)^2+1\): \(4=- (0 + 5)^2+1=-24\), false.

Step4: Check (-5, 1)

Substitute \(x=-5\), \(y = 1\) into \(y=-5x + 4\): \(1=-5(-5)+4=29\), false.

Step5: Analyze intersection

The linear equation \(y=-5x + 4\) (a line) and quadratic \(y=-(x + 5)^2+1\) (a downward - opening parabola with vertex \((-5,1)\)):

• The line has slope \(-5\), y - intercept \(4\).
• The parabola vertex is \((-5,1)\), and it opens down.

By graphing (or solving \(-5x + 4=-(x + 5)^2+1\) → \(x^{2}+10x + 25-5x + 3 = 0\) → \(x^{2}+5x + 28 = 0\), discriminant \(25-112=-87<0\)), so no real solutions.

Answer:

There are no solutions.