Quadratic Functions • Topic 5 of 5

Graph Interpretation

Reading a parabola's graph or equation, the x-intercepts (roots) are where y = 0, the y-intercept is (0, c), and the vertex is the turning point on the axis of symmetry. The factored form y = a(x − r)(x − s) shows the roots r and s immediately. The number of x-intercepts matches the discriminant: two if positive, one if zero, none if negative. Connecting these features — intercepts, vertex, direction — lets you match an equation to its graph and answer "where does it cross," "what is the vertex," and "how many real solutions," all staple SAT graph questions.

✅ Solved examples

1. Where does y = (x − 2)(x − 6) cross the x-axis?
Set each factor to 0: x = 2 and x = 6.
2. How many x-intercepts does y = x² + 1 have?
x² + 1 = 0 has no real solution, so none.
3. What is the y-intercept of y = (x − 1)(x − 4)?
At x = 0, y = (−1)(−4) = 4.
4. A parabola crosses at x = −3 and x = 5. Where is its vertex (x-value)?
Axis x = (−3 + 5)/2 = 1, so the vertex x-coordinate is 1.

✏️ Practice — try these, take hints as needed

1. Where does y = (x + 1)(x − 5) cross the x-axis?
Set each factor to 0.
x + 1 = 0 and x − 5 = 0.
—
x = −1 and x = 5.
2. How many x-intercepts does y = x² − 6x + 9 have?
Discriminant 36 − 36 = 0.
D = 0 → …
—
One.
3. What is the y-intercept of y = (x − 2)(x − 3)?
Set x = 0.
(−2)(−3).
—
6.
4. A parabola crosses at x = 1 and x = 9. Vertex x-value?
Midpoint of the roots.
(1 + 9)/2.
—
5.
5. How many x-intercepts does y = x² + 4 have?
x² + 4 = 0 → x² = −4.
No real square root.
—
None.

📝 Topic test — 8 questions

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