This worksheet provides three practice problems for students to solve systems of linear and quadratic equations by plotting them on the provided coordinate grids.

Let’s solve each system of equations by graphing. We’ll find where the line and parabola intersect — that point (or points) is the solution.

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Problem 1:

Equations:
- y = x - 3 → This is a straight line.
- y = x² - 8x + 15 → This is a parabola.

Step 1: Graph the line y = x - 3.
- When x = 0, y = -3 → point (0, -3)
- When x = 3, y = 0 → point (3, 0)
- When x = 6, y = 3 → point (6, 3)

Draw a straight line through these points.

Step 2: Graph the parabola y = x² - 8x + 15.
First, find vertex: x = -b/(2a) = 8/2 = 4
Then y = (4)² - 8(4) + 15 = 16 - 32 + 15 = -1 → vertex at (4, -1)

Find more points:
- x = 3 → y = 9 - 24 + 15 = 0 → (3, 0)
- x = 5 → y = 25 - 40 + 15 = 0 → (5, 0)
- x = 2 → y = 4 - 16 + 15 = 3 → (2, 3)
- x = 6 → y = 36 - 48 + 15 = 3 → (6, 3)

Plot these and draw the U-shaped curve.

Step 3: Find intersection points.
Looking at the graphs:
- At x = 3, both give y = 0 → (3, 0)
- At x = 6, both give y = 3 → (6, 3)

✔ So solutions are (3, 0) and (6, 3)

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Problem 2:

Equations:
- y = x + 3 → straight line
- y = -x² + 2x + 5 → downward-opening parabola

Step 1: Graph the line y = x + 3.
- x = 0 → y = 3 → (0, 3)
- x = -3 → y = 0 → (-3, 0)
- x = 1 → y = 4 → (1, 4)

Step 2: Graph the parabola y = -x² + 2x + 5
Vertex: x = -b/(2a) = -2/(2*(-1)) = 1
y = -(1)² + 2(1) + 5 = -1 + 2 + 5 = 6 → vertex (1, 6)

More points:
- x = 0 → y = 5 → (0, 5)
- x = 2 → y = -4 + 4 + 5 = 5 → (2, 5)
- x = -1 → y = -1 -2 + 5 = 2 → (-1, 2)
- x = 3 → y = -9 + 6 + 5 = 2 → (3, 2)

Step 3: Find intersections.
Check if any points match:

Try x = -1:
Line: y = -1 + 3 = 2
Parabola: y = 2 → matches! → (-1, 2)

Try x = 2:
Line: y = 2 + 3 = 5
Parabola: y = 5 → matches! → (2, 5)

✔ Solutions: (-1, 2) and (2, 5)

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Problem 3:

Equations:
- y = ½x → line with slope 1/2
- y = x² + 6x + 8 → upward parabola

Step 1: Graph the line y = ½x
- x = 0 → y = 0 → (0, 0)
- x = 2 → y = 1 → (2, 1)
- x = -2 → y = -1 → (-2, -1)
- x = 4 → y = 2 → (4, 2)

Step 2: Graph parabola y = x² + 6x + 8
Vertex: x = -6/(2*1) = -3
y = (-3)² + 6(-3) + 8 = 9 - 18 + 8 = -1 → vertex (-3, -1)

More points:
- x = -2 → y = 4 -12 + 8 = 0 → (-2, 0)
- x = -4 → y = 16 -24 + 8 = 0 → (-4, 0)
- x = -1 → y = 1 -6 + 8 = 3 → (-1, 3)
- x = 0 → y = 8 → (0, 8)

Wait — let’s check for intersections carefully.

We need to see where ½x = x² + 6x + 8

But since we’re graphing, let’s test integer x-values near where they might cross.

Try x = -2:
Line: y = ½*(-2) = -1
Parabola: y = 0 → not equal

Try x = -4:
Line: y = ½*(-4) = -2
Parabola: y = 0 → no

Try x = -1:
Line: y = -0.5
Parabola: y = 3 → no

Try x = -3:
Line: y = -1.5
Parabola: y = -1 → close but not same

Wait — maybe I made a mistake. Let me solve algebraically to verify what the graph should show.

Set ½x = x² + 6x + 8
Multiply both sides by 2:
x = 2x² + 12x + 16
Bring all to one side:
0 = 2x² + 11x + 16

Discriminant: b² - 4ac = 121 - 128 = -7 → negative? That means NO real solutions?

But that can’t be right — let me double-check the problem.

Wait — looking back at the original image transcription:

In Problem 3, it says:
y = ½x
y = x² + 6x + 8

But when I plug in x = -2 into parabola: (-2)^2 + 6*(-2) + 8 = 4 -12 + 8 = 0
Line at x=-2: y = -1 → not equal

x = -4: parabola = 16 -24 +8 = 0; line = -2 → no

x = -1: parabola = 1 -6 +8 = 3; line = -0.5 → no

x = 0: parabola=8, line=0 → no

x = -3: parabola=9-18+8=-1; line=-1.5 → no

Actually, discriminant was negative → no real intersection? But that seems odd for a graphing worksheet.

Wait — perhaps I misread the equation.

Looking again at user input: “y = x² + 6x + 8” — yes.

But let me try solving:

½x = x² + 6x + 8
→ 0 = x² + 5.5x + 8
Multiply by 2: 0 = 2x² + 11x + 16
Discriminant: 121 - 128 = -7 → indeed no real solution.

That means the line and parabola do NOT intersect.

But that would mean "no solution" — which is possible.

However, let me check if there's a typo in my reading.

Wait — in the original problem statement from the image, maybe it’s y = -x² + 6x + 8? Or different coefficient?

No — according to the text provided: “y = x² + 6x + 8”

Alternatively, perhaps the line is y = -½x? But no, it says y = ½x.

Given the math, there are no real solutions.

But let’s think — maybe I should graph approximate values.

At x = -2: line y=-1, parabola y=0 → parabola above
At x = -3: line y=-1.5, parabola y=-1 → parabola still above
At x = -4: line y=-2, parabola y=0 → parabola way above
At x = 0: line y=0, parabola y=8 → parabola above
At x = -1: line y=-0.5, parabola y=3 → parabola above

The parabola opens up, vertex at (-3,-1), and the line has positive slope passing through origin.

The lowest point of parabola is y=-1 at x=-3, while line at x=-3 is y=-1.5 — so line is below parabola everywhere.

And since parabola goes to infinity as x increases or decreases, and line is straight, they never meet.

So actually — no solution.

But that feels strange for a homework problem. Maybe I made an error.

Wait — let me recalculate discriminant:

Equation: ½x = x² + 6x + 8
Subtract ½x: 0 = x² + 5.5x + 8
Multiply by 2: 0 = 2x² + 11x + 16
Discriminant D = 11² - 4*2*16 = 121 - 128 = -7 < 0 → no real roots.

So mathematically, no intersection.

Therefore, for Problem 3, there is no solution.

But let me confirm with another approach.

Suppose we set f(x) = x² + 6x + 8 - ½x = x² + 5.5x + 8

Minimum value of this quadratic: at x = -5.5/2 = -2.75

f(-2.75) = (-2.75)^2 + 5.5*(-2.75) + 8 = 7.5625 - 15.125 + 8 = 0.4375 > 0

So the difference is always positive → parabola always above line → no intersection.

✔ So Problem 3 has no solution.

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Final Answer:

Problem 1: (3, 0) and (6, 3)
Problem 2: (-1, 2) and (2, 5)
Problem 3: No solution