Vertex Form and Intercept Form Worksheet for Quadratic Functions
A free printable worksheet with a guide to reading, writing, and converting quadratics between vertex form and intercept form
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A quadratic can be written three ways, and each one hands you something different for free. Standard form, y = ax² + bx + c, is the one most students meet first, but it hides the two things you usually want to know: where the parabola turns, and where it crosses the x-axis. Vertex form and intercept form each put one of those front and center.
This guide covers what each form tells you at a glance, how to move a quadratic from one form to another, and where students tend to slip. The worksheet that follows gives you twenty to practice on, with a full answer key.
Vertex form: y = a(x − h)² + k
The vertex is (h, k), read straight off the equation. Watch the sign: the h inside the bracket flips.
Intercept form: y = a(x − p)(x − q)
The x-intercepts are p and q, again with the signs flipped. The vertex sits halfway between them.
Vertex Form: y = a(x − h)² + k
Vertex form is built so the vertex is impossible to miss. The point (h, k) is the turning point of the parabola, and it comes straight out of the equation with no work. The catch is the sign of h. Because the form has a minus sign built in, y = (x − 3)² + 2 has its vertex at x = 3, while y = (x + 3)² + 2 has its vertex at x = −3. The number you see is not the answer until you account for that built-in minus.
The value of a still controls the shape. A positive a opens the parabola upward and makes the vertex a minimum; a negative a opens it downward and makes the vertex a maximum. A larger a makes the curve narrower. None of that changes where the vertex is, only how the curve behaves around it. If you want more practice locating that turning point from standard form, the vertex and axis of symmetry worksheet covers the −b ⁄ 2a route in depth, and Khan Academy walks through the same form from the graphing side.
Intercept Form: y = a(x − p)(x − q)
Intercept form, sometimes called factored form, is built around where the parabola crosses the x-axis. Set each factor equal to zero and you get the two x-intercepts, p and q. The same sign warning applies: y = (x + 1)(x − 3) crosses at x = −1 and x = 3, not at 1 and 3.
This form also gives you the vertex with almost no effort, because a parabola is symmetric. The vertex always sits exactly halfway between the two x-intercepts, so you average p and q to get its x-coordinate, then substitute back to find the y. For the parabola below, the roots are −1 and 3, their midpoint is 1, and one substitution finishes the vertex.
When a student stalls here, it is almost never the factoring. It is that they find the two roots, average them correctly, and then stop — forgetting that the midpoint is only the x-coordinate. The vertex needs a y as well, and that means substituting back into the original equation. I ask students to write the vertex as an ordered pair from the very first line so the empty second slot nags at them until it is filled.
Both Forms Describe the Same Curve
It helps to see that these are not different parabolas, just different ways of writing one. The curve below can be written as 2(x − 1)² − 8 or as 2(x + 1)(x − 3), and both are correct. The vertex form tells you the turning point is (1, −8); the intercept form tells you it crosses at −1 and 3. Multiply either one out and you land on the same standard form, y = 2x² − 4x − 6.
Moving Between the Forms
Converting is where most of the actual work lives. To go from intercept form to standard form, you multiply the two factors out. To go from vertex form to standard form, you expand the square and simplify. Both directions are just careful algebra.
Going the other way is harder. To reach intercept form from standard form you factor the quadratic, which only works cleanly when the roots are nice numbers, so solving quadratics by factoring is the skill this depends on. To reach vertex form you complete the square, which works every time but takes more steps. When a quadratic refuses to factor, the quadratic formula will still find the roots, and the discriminant tells you how many to expect before you start. The OpenStax Algebra and Trigonometry chapter sets out both forms if you want a textbook treatment.
Worked Example: standard form to vertex form
y = x² − 6x + 5
Start by taking half of the middle coefficient and squaring it. Half of −6 is −3, and (−3)² is 9. Add and subtract that 9 inside the expression: y = x² − 6x + 9 − 9 + 5. The first three terms are now a perfect square, (x − 3)², and the leftover constants combine to −4. That gives y = (x − 3)² − 4.
Vertex form: y = (x − 3)² − 4, so the vertex is (3, −4). Check it against intercept form: x² − 6x + 5 factors to (x − 1)(x − 5), roots 1 and 5, whose midpoint is 3. Same vertex x, as it must be.
Worked Example: standard form to intercept form
y = x² − 2x − 3
This one factors. You want two numbers that multiply to −3 and add to −2, which are −3 and 1. That gives y = (x − 3)(x + 1), so the roots are 3 and −1. Averaging them puts the vertex at x = 1, and substituting back gives y = −4.
Intercept form: y = (x − 3)(x + 1), roots at x = 3 and x = −1. The factor (x + 1) gives a root of −1, not +1. Ten years and six thousand hours in, this is still the single mistake I correct more than any other in this topic — students read the root off the factor without flipping the sign.
Practice Problems: Vertex Form and Intercept Form Worksheet
Part A gives you quadratics in standard form to rewrite in intercept form, then read off the roots and vertex. Part B asks for vertex form by completing the square. Show your work in the space provided. A few equations appear in both parts on purpose — writing the same curve two ways is the point of the exercise.
Part A — write in intercept form, then give the roots and vertex
- y = x² − 4x − 5
- y = x² + 4x + 3
- y = x² − 6x + 8
- y = x² + 2x − 15
- y = x² − 8x + 12
- y = x² + 6x + 5
- y = x² − 2x − 8
- y = x² − 9
- y = 2x² − 4x − 6
- y = x² − 5x + 6
Part B — write in vertex form by completing the square, then give the vertex
- y = x² − 4x + 1
- y = x² + 6x + 5
- y = x² − 2x − 8
- y = x² + 8x + 10
- y = x² − 10x + 21
- y = x² + 2x − 5
- y = x² − 6x + 4
- y = x² + 4x + 9
- y = 2x² − 8x + 3
- y = x² − 12x + 40
Worksheet: Answer Key
Every problem from the vertex form and intercept form worksheet is worked out below. Click a question to reveal the full solution.
Part A — intercept form
1. y = x² − 4x − 5
Factors to (x − 5)(x + 1). Roots 5 and −1, midpoint 2, y = −9.y = (x − 5)(x + 1) · Roots 5 and −1 · Vertex (2, −9)
2. y = x² + 4x + 3
Factors to (x + 1)(x + 3). Roots −1 and −3, midpoint −2, y = −1.y = (x + 1)(x + 3) · Roots −1 and −3 · Vertex (−2, −1)
3. y = x² − 6x + 8
Factors to (x − 2)(x − 4). Roots 2 and 4, midpoint 3, y = −1.y = (x − 2)(x − 4) · Roots 2 and 4 · Vertex (3, −1)
4. y = x² + 2x − 15
Factors to (x + 5)(x − 3). Roots −5 and 3, midpoint −1, y = −16.y = (x + 5)(x − 3) · Roots −5 and 3 · Vertex (−1, −16)
5. y = x² − 8x + 12
Factors to (x − 2)(x − 6). Roots 2 and 6, midpoint 4, y = −4.y = (x − 2)(x − 6) · Roots 2 and 6 · Vertex (4, −4)
6. y = x² + 6x + 5
Factors to (x + 1)(x + 5). Roots −1 and −5, midpoint −3, y = −4.y = (x + 1)(x + 5) · Roots −1 and −5 · Vertex (−3, −4)
7. y = x² − 2x − 8
Factors to (x − 4)(x + 2). Roots 4 and −2, midpoint 1, y = −9.y = (x − 4)(x + 2) · Roots 4 and −2 · Vertex (1, −9)
8. y = x² − 9
Difference of squares: (x − 3)(x + 3). Roots 3 and −3, midpoint 0, y = −9.y = (x − 3)(x + 3) · Roots 3 and −3 · Vertex (0, −9)
9. y = 2x² − 4x − 6
Factor out 2 first: 2(x² − 2x − 3) = 2(x − 3)(x + 1). Roots 3 and −1, midpoint 1, y = −8.y = 2(x − 3)(x + 1) · Roots 3 and −1 · Vertex (1, −8)
10. y = x² − 5x + 6
Factors to (x − 2)(x − 3). Roots 2 and 3, midpoint 5/2, y = −1/4.y = (x − 2)(x − 3) · Roots 2 and 3 · Vertex (5/2, −1/4)
Part B — vertex form (completing the square)
11. y = x² − 4x + 1
Half of −4 is −2, squared is 4. y = (x² − 4x + 4) − 4 + 1 = (x − 2)² − 3.y = (x − 2)² − 3 · Vertex (2, −3)
12. y = x² + 6x + 5
Half of 6 is 3, squared is 9. y = (x² + 6x + 9) − 9 + 5 = (x + 3)² − 4.y = (x + 3)² − 4 · Vertex (−3, −4)
13. y = x² − 2x − 8
Half of −2 is −1, squared is 1. y = (x² − 2x + 1) − 1 − 8 = (x − 1)² − 9.y = (x − 1)² − 9 · Vertex (1, −9)
14. y = x² + 8x + 10
Half of 8 is 4, squared is 16. y = (x² + 8x + 16) − 16 + 10 = (x + 4)² − 6.y = (x + 4)² − 6 · Vertex (−4, −6)
15. y = x² − 10x + 21
Half of −10 is −5, squared is 25. y = (x² − 10x + 25) − 25 + 21 = (x − 5)² − 4.y = (x − 5)² − 4 · Vertex (5, −4)
16. y = x² + 2x − 5
Half of 2 is 1, squared is 1. y = (x² + 2x + 1) − 1 − 5 = (x + 1)² − 6.y = (x + 1)² − 6 · Vertex (−1, −6)
17. y = x² − 6x + 4
Half of −6 is −3, squared is 9. y = (x² − 6x + 9) − 9 + 4 = (x − 3)² − 5.y = (x − 3)² − 5 · Vertex (3, −5)
18. y = x² + 4x + 9
Half of 4 is 2, squared is 4. y = (x² + 4x + 4) − 4 + 9 = (x + 2)² + 5.y = (x + 2)² + 5 · Vertex (−2, 5)
19. y = 2x² − 8x + 3
Factor 2 from the x-terms: 2(x² − 4x) + 3. Complete inside: 2(x² − 4x + 4 − 4) + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5.y = 2(x − 2)² − 5 · Vertex (2, −5)
20. y = x² − 12x + 40
Half of −12 is −6, squared is 36. y = (x² − 12x + 36) − 36 + 40 = (x − 6)² + 4.y = (x − 6)² + 4 · Vertex (6, 4)
Common Questions
What is the difference between vertex form and intercept form?
Vertex form, y = a(x − h)² + k, hands you the turning point (h, k) directly. Intercept form, y = a(x − p)(x − q), hands you the two x-intercepts, p and q. Both describe the same parabola and both expand to the same standard form.
Is intercept form the same as factored form?
Yes. Teachers and textbooks use both names for y = a(x − p)(x − q). "Factored form" describes how it was written; "intercept form" describes what it tells you. If your class uses one term and your worksheet uses the other, they are the same thing.
Can every quadratic be written in intercept form?
No. Intercept form needs real x-intercepts, so it only exists when the parabola actually crosses or touches the x-axis. Check the discriminant first: if b² − 4ac is negative there are no real roots and no intercept form, though vertex form still works.
How do you find the vertex from intercept form?
Average the two roots to get the x-coordinate, since a parabola is symmetric about its vertex. Then substitute that x back into the equation to get the y. For y = (x + 1)(x − 3), the roots are −1 and 3, the midpoint is 1, and substituting gives y = −4.
Which form should you use for graphing?
Whichever matches what the question asks for. If you need the maximum or minimum, use vertex form. If you need the roots or a rough sketch of where the curve sits, use intercept form. Most graphing questions are fastest when you get both and plot the vertex, the two roots, and the y-intercept.
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