Question

on the set of axes below, solve the following system of equations graphically and state the coordinates of all points in the solution set.
$y = x^2 + 4x + 4$
$y = 2x + 7$
you can move the parabola by dragging the dots. graph the line by clicking twice.

Explanation:

Step1: Set the equations equal

Since both equal \( y \), set \( x^2 + 4x + 4 = 2x + 7 \).

Step2: Simplify the equation

Rearrange to \( x^2 + 2x - 3 = 0 \).

Step3: Factor the quadratic

Factor to \( (x + 3)(x - 1) = 0 \).

Step4: Solve for \( x \)

Set each factor to zero: \( x + 3 = 0 \) gives \( x = -3 \); \( x - 1 = 0 \) gives \( x = 1 \).

Step5: Find corresponding \( y \)

For \( x = -3 \), \( y = 2(-3) + 7 = 1 \). For \( x = 1 \), \( y = 2(1) + 7 = 9 \). Wait, but the parabola \( y = x^2 + 4x + 4 = (x + 2)^2 \), vertex at \( (-2, 0) \). Wait, maybe graphically: the parabola \( y=(x + 2)^2 \) (vertex at \( (-2, 0) \)), line \( y = 2x + 7 \). Wait, earlier algebra had a mistake. Let's redo: \( x^2 + 4x + 4 = 2x + 7 \) → \( x^2 + 2x - 3 = 0 \). Wait, discriminant \( 4 + 12 = 16 \), roots \( \frac{-2 \pm 4}{2} \), so \( x = 1 \) or \( x = -3 \). Then \( y \) for \( x = -3 \): \( (-3)^2 + 4(-3) + 4 = 9 - 12 + 4 = 1 \). For \( x = 1 \): \( 1 + 4 + 4 = 9 \). But the graph in the image has the parabola vertex at (0,0)? Wait, maybe the given parabola in the image is misdrawn, but following the equations: the system is \( y = x^2 + 4x + 4 \) (which is \( (x + 2)^2 \), vertex at (-2, 0)) and \( y = 2x + 7 \). The solutions are the intersection points. But maybe the image's parabola is a different one, but algebraically, solving \( x^2 + 4x + 4 = 2x + 7 \) gives \( x = -3, y = 1 \) and \( x = 1, y = 9 \). Wait, but maybe the problem's graph has a different parabola. Wait, the user's graph shows a parabola with vertex at (0,0), maybe the equation was mistyped? If the parabola is \( y = x^2 \), then \( x^2 = 2x + 7 \) → \( x^2 - 2x - 7 = 0 \), roots \( 1 \pm 2\sqrt{2} \). But the original problem says \( y = x^2 + 4x + 4 \). There's confusion. Wait, the image's parabola has vertex at (0,0) (green dot) and a blue dot at (1, -1)? No, axes are flipped: y-axis is horizontal, x-axis vertical? Wait, the axes are labeled with y horizontal (right) and x vertical (down)? That's a rotated coordinate system. So in the image, horizontal axis is y, vertical is x. So the parabola is \( x = y^2 + 4y + 4 \) (since x is vertical), and the line is \( x = 2y + 7 \). Then set \( y^2 + 4y + 4 = 2y + 7 \) → \( y^2 + 2y - 3 = 0 \) → \( (y + 3)(y - 1) = 0 \), so \( y = -3 \) or \( y = 1 \). Then \( x = 2(-3) + 7 = 1 \), \( x = 2(1) + 7 = 9 \). So the points are (x, y) = (1, -3) and (9, 1) in standard coordinates, but in the image's rotated axes (y horizontal, x vertical), the coordinates would be (y, x) = (-3, 1) and (1, 9). Looking at the graph, the green dot is at (0, 0) (y=0, x=0), blue dot at (1, 1) (y=1, x=1). Wait, maybe the problem's equations are in the rotated axes. So solving the system with rotated axes (y horizontal, x vertical): \( x = y^2 + 4y + 4 \) and \( x = 2y + 7 \). Solving gives \( y = -3 \), \( x = 1 \) and \( y = 1 \), \( x = 9 \). So the solution set is the points where they intersect, which are (x, y) in rotated axes: when y = -3, x = 1; y = 1, x = 9. But in the image, the parabola passes through (0,0) (y=0, x=0: 0 = 0 + 0 + 4? No, so maybe the parabola equation is \( x = y^2 \) (vertex at (0,0)), then \( y^2 = 2y + 7 \) → \( y^2 - 2y - 7 = 0 \), roots \( 1 \pm 2\sqrt{2} \), which doesn't match. There's a discrepancy, but following the given equations \( y = x^2 + 4x + 4 \) and \( y = 2x + 7 \) (standard axes), the solutions are (-3, 1) and (1, 9). But if axes are rotated (y horizontal, x vertical), then the solutions are (1, -3) and (9, 1).

Answer:

The solution set is the points of intersection. Solving algebraically (correcting axis confusion or not), the solutions are \(\boldsymbol{(-3, 1)}\) and \(\boldsymbol{(1, 9)}\) (in standard \( (x, y) \) coordinates) or, if axes are rotated (y horizontal, x vertical), \((1, -3)\) and \((9, 1)\). But likely, with standard axes, the solutions are \((-3, 1)\) and \((1, 9)\). Wait, no—wait \( y = x^2 + 4x + 4 \) at \( x = -3 \): \( (-3)^2 + 4(-3) + 4 = 9 - 12 + 4 = 1 \), correct. At \( x = 1 \): \( 1 + 4 + 4 = 9 \), correct. So the solution set is \(\{ (-3, 1), (1, 9) \}\).