Identify Polynomial Functions worksheet with equations and graph options for students to match.

Let's solve each problem step by step and identify the correct graph for each given polynomial function.

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Problem 1: Identify the graph of $ y = 2x - 2 $

This is a linear function (degree 1), so its graph is a straight line.

#### Step 1: Find two points on the line
We can plug in values of $ x $ to find corresponding $ y $-values.

- When $ x = 0 $:
$ y = 2(0) - 2 = -2 $ → Point: $ (0, -2) $
- When $ x = 1 $:
$ y = 2(1) - 2 = 0 $ → Point: $ (1, 0) $
- When $ x = 2 $:
$ y = 2(2) - 2 = 2 $ → Point: $ (2, 2) $

So the line passes through $ (0, -2) $, $ (1, 0) $, and $ (2, 2) $. It has a positive slope (since coefficient of $ x $ is +2).

Now look at the graphs:

- A) A straight line passing through $ (0, -2) $, going up to the right — looks like it goes through $ (1, 0) $, $ (2, 2) $ — ✔ Correct
- B) Also a straight line, but seems to pass through $ (0, 0) $ or close — not $ (0, -2) $ ✘
- C) Parabola — quadratic — ✘
- D) Line with positive slope, but appears to go through $ (0, 0) $ — not $ (0, -2) $ ✘

✔ So, Answer for #1: A

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Problem 2: Identify the graph of $ y = 3x^2 - 2x + 1 $

This is a quadratic function (degree 2), so its graph is a parabola.

#### Step 1: Determine direction of opening
Coefficient of $ x^2 $ is $ 3 > 0 $ → parabola opens upward

So eliminate any graphs that open downward:
- C) Opens downward ✘
- A) Looks like a cubic (odd degree, ends go opposite directions) ✘
- B) Opens upward — possible ✔
- D) Opens upward — possible ✔

Now we need to distinguish between B and D.

#### Step 2: Check the vertex and shape

The vertex form helps, but let’s find the vertex:

Vertex formula: $ x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3} $

Now find $ y $ at $ x = \frac{1}{3} $:

$$
y = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) + 1 = 3\left(\frac{1}{9}\right) - \frac{2}{3} + 1 = \frac{1}{3} - \frac{2}{3} + 1 = -\frac{1}{3} + 1 = \frac{2}{3}
$$

So the vertex is at $ \left(\frac{1}{3}, \frac{2}{3}\right) $ — which is approximately $ (0.33, 0.67) $

That means:
- The vertex is to the right of the y-axis, and above the x-axis
- Since the minimum value is $ \frac{2}{3} > 0 $, the entire parabola is above the x-axis — no x-intercepts

Now examine graphs:

- B) Parabola opening upward, vertex near $ (1, -1) $? Wait — the lowest point seems below the x-axis → would have real roots → but our function has no real roots (discriminant negative) → ✘
- D) Vertex is around $ (0.3, 0.7) $, opens upward, stays above x-axis → ✔

Check discriminant to confirm:
$$
\Delta = b^2 - 4ac = (-2)^2 - 4(3)(1) = 4 - 12 = -8 < 0
$$
No real roots → never crosses x-axis → always positive → matches Graph D

Also, Graph D has vertex slightly to the right of origin and above x-axis — matches.

✔ So, Answer for #2: D

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✔ Final Answers:

1) $ y = 2x - 2 $ → A
2) $ y = 3x^2 - 2x + 1 $ → D

Let me know if you'd like a visual explanation or more detail!