Math worksheet illustrating how to solve systems of linear and quadratic equations through graphing and substitution.

Let's solve the problems in your worksheet step by step. We'll analyze each example, determine the number of solutions (based on intersections), and find the actual solutions where possible.

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Part 1: Types of Solutions Produced by a Linear and Quadratic System

We are given three examples with graphs. The goal is to determine:

- Number of solutions (points of intersection)
- What are the solutions? (coordinates of intersection points)

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#### Example 1:
$$
\begin{cases}
y = x^2 - 2x - 3 \\
y = 2x - 3
\end{cases}
$$

Graph Analysis:
- The blue curve is a parabola opening upwards.
- The red line is straight.
- They intersect at two points.

From the graph:
- One point appears near $(-1, -5)$
- Another near $(3, 3)$

Let’s verify algebraically:

Set equations equal:
$$
x^2 - 2x - 3 = 2x - 3
$$
$$
x^2 - 4x = 0 \Rightarrow x(x - 4) = 0
\Rightarrow x = 0 \text{ or } x = 4
$$

Wait — this contradicts the graph. Let's check again.

Wait! Graph shows intersection at $x = -1$ and $x = 3$. Let's plug into both equations.

Try $x = -1$:
- $y = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$
- $y = 2(-1) - 3 = -2 - 3 = -5$ → Not matching!

Wait — there might be an error in reading the graph.

Let’s re-solve carefully.

Set:
$$
x^2 - 2x - 3 = 2x - 3
\Rightarrow x^2 - 4x = 0 \Rightarrow x(x - 4) = 0
\Rightarrow x = 0 \text{ or } x = 4
$$

Now find $y$:
- If $x = 0$, $y = 2(0) - 3 = -3$
- If $x = 4$, $y = 2(4) - 3 = 8 - 3 = 5$

So solutions: $(0, -3)$ and $(4, 5)$

But graph seems to show different points?

Wait — let's double-check the graph. The parabola $y = x^2 - 2x - 3$ has vertex at $x = 1$, $y = 1 - 2 - 3 = -4$, so vertex $(1, -4)$

At $x = 0$: $y = 0 - 0 - 3 = -3$

At $x = 4$: $y = 16 - 8 - 3 = 5$

And line $y = 2x - 3$: at $x=0$, $y=-3$; $x=4$, $y=5$

So yes, they pass through $(0, -3)$ and $(4, 5)$

But graph looks like it intersects at $(-1, -5)$ and $(3, 3)$? That doesn’t match.

Wait — perhaps the graph is misleading. Let's trust the math.

But looking at the graph: the red line goes from bottom-left to top-right, crosses parabola at two points.

At $x = 0$, both give $y = -3$ → point $(0, -3)$

At $x = 4$, both give $y = 5$ → point $(4, 5)$

So the graph must be inaccurate, or we're misreading.

But wait — maybe the graph is correct and I made a mistake.

Wait — let's plot:

Parabola: $y = x^2 - 2x - 3$

Factor: $y = (x - 3)(x + 1)$ → x-intercepts at $x = -1, 3$

Vertex: $x = 1$, $y = 1 - 2 - 3 = -4$

So vertex at $(1, -4)$

Line: $y = 2x - 3$

At $x = -1$: $y = -2 - 3 = -5$ → point $(-1, -5)$

Check if on parabola: $(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0 ≠ -5$ → not on parabola

But at $x = -1$, parabola gives $y = 0$, line gives $y = -5$ → no

Wait — but on graph, they appear to intersect at $(-1, -5)$?

No — that can't be.

Wait — perhaps the line is not $y = 2x - 3$?

Wait — let's look back.

The system is:
$$
y = x^2 - 2x - 3 \\
y = 2x - 3
$$

Set equal:
$$
x^2 - 2x - 3 = 2x - 3 \\
x^2 - 4x = 0 \\
x(x - 4) = 0 \Rightarrow x = 0, x = 4
$$

So only two solutions: $(0, -3)$ and $(4, 5)$

So number of solutions: 2

Solutions: $(0, -3)$ and $(4, 5)$

But graph shows something else? Maybe the graph is wrong or mislabeled.

Alternatively, perhaps the line is different.

Wait — maybe the line is $y = 2x + 3$? No, it says $y = 2x - 3$

Let me check the graph visually.

In Example 1, the red line passes through $(0, -3)$ and $(4, 5)$ — yes, slope = $(5+3)/(4-0) = 8/4 = 2$, so $y = 2x - 3$ is correct.

And parabola passes through $(-1, 0)$, $(0, -3)$, $(1, -4)$, $(3, 0)$, $(4, 5)$

At $x = 0$: both give $y = -3$ → intersection

At $x = 4$: both give $y = 5$ → second intersection

So the graph should show intersection at $(0, -3)$ and $(4, 5)$

But visually, it may seem like more, but likely only two.

So:

✔ Number of solutions: 2

✔ Solutions: (0, -3) and (4, 5)

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#### Example 2:
$$
\begin{cases}
y = -(x + 2)^2 + 5 \\
y = 5
\end{cases}
$$

This is a downward-opening parabola with vertex at $(-2, 5)$

The line $y = 5$ is horizontal.

Since the vertex is at $y = 5$, and the parabola opens down, it touches the line $y = 5$ only at the vertex.

So: one solution

At $x = -2$, $y = 5$

So:

✔ Number of solutions: 1

✔ Solution: (-2, 5)

---

#### Example 3:
$$
\begin{cases}
y = x^2 - 2x + 4 \\
y = x - 1
\end{cases}
$$

Parabola: $y = x^2 - 2x + 4$, opens up.

Vertex: $x = 1$, $y = 1 - 2 + 4 = 3$ → vertex $(1, 3)$

Line: $y = x - 1$

Set equal:
$$
x^2 - 2x + 4 = x - 1 \\
x^2 - 3x + 5 = 0
$$

Discriminant: $(-3)^2 - 4(1)(5) = 9 - 20 = -11 < 0$

No real solutions.

So the line does not intersect the parabola.

From the graph: the parabola is above the line everywhere.

✔ Number of solutions: 0

✔ Solutions: None (no real solutions)

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✔ Summary for Part 1:

| Example | Number of Solutions | Solutions |
|--------|---------------------|-----------|
| 1 | 2 | $(0, -3)$, $(4, 5)$ |
| 2 | 1 | $(-2, 5)$ |
| 3 | 0 | None |

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Part 2: Solving Linear and Quadratic Systems by Graphing

We’ll now solve Examples 4a, 5a, 6a by graphing (but since we can't draw here, we’ll find the solutions algebraically and describe how to graph).

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#### Example 4a:
$$
\begin{cases}
y = (x + 2)^2 - 6 \\
y = 4x - 2
\end{cases}
$$

First, expand:
$$
y = (x + 2)^2 - 6 = x^2 + 4x + 4 - 6 = x^2 + 4x - 2
$$

Set equal:
$$
x^2 + 4x - 2 = 4x - 2 \\
x^2 = 0 \Rightarrow x = 0
$$

Then $y = 4(0) - 2 = -2$

So only one solution: $(0, -2)$

Graphing: Parabola $y = (x+2)^2 - 6$ has vertex at $(-2, -6)$, opens up. Line $y = 4x - 2$ passes through $(0, -2)$, $(1, 2)$, etc.

They touch at $(0, -2)$ — tangent.

✔ Number of solutions: 1

✔ Solution: (0, -2)

---

#### Example 5a:
$$
\begin{cases}
y = x^2 - 2x - 3 \\
y = -5
\end{cases}
$$

Set equal:
$$
x^2 - 2x - 3 = -5 \\
x^2 - 2x + 2 = 0
$$

Discriminant: $(-2)^2 - 4(1)(2) = 4 - 8 = -4 < 0$

No real solutions.

But let's see: parabola $y = x^2 - 2x - 3$, minimum at $x = 1$, $y = 1 - 2 - 3 = -4$

So minimum value is $-4$, and $y = -5$ is below it → no intersection.

✔ Number of solutions: 0

✔ Solutions: None

---

#### Example 6a:
$$
\begin{cases}
y = -x^2 + 2x + 7 \\
y = -2x + 2
\end{cases}
$$

Set equal:
$$
-x^2 + 2x + 7 = -2x + 2 \\
-x^2 + 4x + 5 = 0 \\
Multiply by -1: x^2 - 4x - 5 = 0 \\
Factor: (x - 5)(x + 1) = 0 \\
x = 5 \text{ or } x = -1
$$

Find $y$:

- If $x = 5$, $y = -2(5) + 2 = -10 + 2 = -8$
- If $x = -1$, $y = -2(-1) + 2 = 2 + 2 = 4$

So solutions: $(-1, 4)$ and $(5, -8)$

✔ Number of solutions: 2

✔ Solutions: (-1, 4), (5, -8)

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✔ Summary for Part 2:

| Example | Number of Solutions | Solutions |
|--------|---------------------|-----------|
| 4a | 1 | $(0, -2)$ |
| 5a | 0 | None |
| 6a | 2 | $(-1, 4)$, $(5, -8)$ |

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Part 3: Solving by Substitution (Examples 4b, 5b, 6b)

These are likely the same systems as 4a, 5a, 6a, but solved algebraically.

Let’s do them fully.

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#### Example 4b:
Same as 4a:
$$
y = (x + 2)^2 - 6 = x^2 + 4x + 4 - 6 = x^2 + 4x - 2 \\
y = 4x - 2
$$

Set equal:
$$
x^2 + 4x - 2 = 4x - 2 \\
x^2 = 0 \Rightarrow x = 0 \\
y = 4(0) - 2 = -2
$$

✔ Solution: $(0, -2)$

---

#### Example 5b:
$$
y = x^2 - 2x - 3 \\
y = -5
$$

Set equal:
$$
x^2 - 2x - 3 = -5 \\
x^2 - 2x + 2 = 0 \\
\text{Discriminant: } 4 - 8 = -4 < 0
\Rightarrow \text{No real solutions}
$$

✔ No real solutions

---

#### Example 6b:
$$
y = -x^2 + 2x + 7 \\
y = -2x + 2
$$

Set equal:
$$
-x^2 + 2x + 7 = -2x + 2 \\
-x^2 + 4x + 5 = 0 \\
x^2 - 4x - 5 = 0 \\
(x - 5)(x + 1) = 0 \Rightarrow x = 5, x = -1
$$

Then:
- $x = 5$: $y = -2(5) + 2 = -8$
- $x = -1$: $y = -2(-1) + 2 = 4$

✔ Solutions: $(-1, 4)$, $(5, -8)$

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✔ Final Answers:

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Types of Solutions Produced By a Linear and Quadratic System

| Example | Number of Solutions | Solutions |
|--------|---------------------|-----------|
| 1 | 2 | $(0, -3)$, $(4, 5)$ |
| 2 | 1 | $(-2, 5)$ |
| 3 | 0 | None |

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Solving by Graphing

| Example | Number of Solutions | Solutions |
|--------|---------------------|-----------|
| 4a | 1 | $(0, -2)$ |
| 5a | 0 | None |
| 6a | 2 | $(-1, 4)$, $(5, -8)$ |

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Solving by Substitution

| Example | Solutions |
|--------|-----------|
| 4b | $(0, -2)$ |
| 5b | No real solutions |
| 6b | $(-1, 4)$, $(5, -8)$ |

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🔍 Key Concept:

A system of a linear and quadratic equation can have:
- 0 solutions: line doesn't intersect parabola
- 1 solution: line is tangent to parabola
- 2 solutions: line crosses parabola twice

This depends on the discriminant of the resulting quadratic equation after substitution.

Let me know if you'd like help sketching any of the graphs!