SAT Vertex Form on Desmos: Quadratic Problems Solved

In vertex form, y = a(x - h)² + k, the vertex is (h, k), so y = 2(x - 3)² - 5 has its vertex at (3, -5). To get a quadratic into vertex form, complete the square. Then graph both versions on Desmos and check that they match. You can type every Desmos line on this page into the College Board version of Desmos on our Desmos on the SAT page.

Bluebook has a Desmos calculator built in, and College Board allows calculators on the whole Math section. This page works four SAT-style vertex form problems, then gives you three practice problems with answers. If your quadratic comes in factored form and you need its lowest point, read how to find a quadratic's minimum on Desmos instead.

How to read vertex form

Vertex form is y = a(x - h)² + k. You can read three facts straight off it.

• The vertex is (h, k). The sign inside the parentheses flips. In y = (x + 2)² + 5, x + 2 means x - (-2), so h = -2 and the vertex is (-2, 5).
• The axis of symmetry is x = h. Any two x-values the same distance from h give the same y-value.
• The sign of a sets the direction. If a > 0, the parabola opens up and k is the minimum value. If a < 0, it opens down and k is the maximum value.

How to convert standard form to vertex form

Complete the square. Take y = x² + 8x + 10.

1. Take half of the x coefficient: half of 8 is 4. Square it: 16.
2. Add and subtract 16: y = (x² + 8x + 16) - 16 + 10.
3. Write the square and combine the numbers: y = (x + 4)² - 6.

The vertex is (-4, -6). When a is not 1, factor a out of the x terms first. Problem 1 shows that case.

To find only the vertex, you can skip the conversion. In y = ax² + bx + c, the vertex sits at x = -b/(2a). Substitute that x to get the y-value.

How to check your vertex on Desmos

Desmos's help center says that when you click a line or curve, gray dots appear at intercepts, intersection points, maximums and minimums. Click a gray dot to see its coordinates. That gives you two checks on any conversion:

1  y = x^2 + 8x + 10
2  y = (x + 4)^2 - 6

If your vertex form is right, line 2 lands on line 1 and you see one parabola. If you see two, your conversion has an error. Then click the curve and click the gray dot at the bottom. It should read (-4, -6).

Problem 1: standard form to vertex form (I do)

The function g is defined by g(x) = 2x² - 12x + 13. The graph of y = g(x) in the xy-plane has its vertex at (h, k). What is the value of k?

Complete the square. Here a = 2, so factor 2 out of the x terms first:

1. g(x) = 2(x² - 6x) + 13
2. Half of -6 is -3, and (-3)² = 9. Add and subtract 9 inside: g(x) = 2(x² - 6x + 9 - 9) + 13.
3. Move the -9 out. It is inside the 2, so it leaves as 2(-9) = -18: g(x) = 2(x² - 6x + 9) - 18 + 13.
4. g(x) = 2(x - 3)² - 5.

The vertex is (3, -5), so k = -5. Since a = 2 > 0, the parabola opens up and -5 is the minimum value.

Check on Desmos:

1  g(x) = 2x^2 - 12x + 13
2  y = 2(x - 3)^2 - 5
3  g(3)

Line 2 lands on line 1, and line 3 returns -5. Click the curve and the gray dot at its lowest point reads (3, -5).

Check by hand. x = -b/(2a) = 12/4 = 3, and g(3) = 2(9) - 36 + 13 = -5.

Problem 2: build the equation from a vertex and a point (we do)

A parabola has its vertex at (-2, 5) and passes through the point (0, 1). Which equation represents the parabola?

A) y = (x + 2)² + 5
B) y = -(x + 2)² + 5
C) y = 2(x + 2)² + 5
D) y = -(x - 2)² + 5

Step 1 (shown). Put the vertex into vertex form. h = -2 and k = 5, so y = a(x + 2)² + 5. Choice D uses (x - 2), which puts the vertex at (2, 5), so cross it out.

Step 2 (shown). Substitute the point (0, 1): 1 = a(0 + 2)² + 5, so 1 = 4a + 5.

Step 3 (you finish). Solve for a and pick the choice. Then check it on Desmos:

1  f(x) = -(x + 2)^2 + 5
2  f(0)

Check your answer: 4a = -4, so a = -1 and the answer is B, y = -(x + 2)² + 5. Line 2 returns 1, and the top gray dot on the curve reads (-2, 5). In standard form this is y = -x² - 4x + 1. Watch choice D: it also passes through (0, 1), because -(0 - 2)² + 5 = 1. Only the vertex rules it out.

Problem 3: a symmetry condition and a "must be true" question

The function f is defined by f(x) = ax² - 7x + c, where a and c are constants and c > 0. The graph of y = f(x) is a parabola with vertex (h, k), and f(-9) = f(-2). Which of the following must be true?

I. a < -3/4
II. k > 16

A) I only
B) II only
C) I and II
D) Neither I nor II

Find h from the symmetry. f(-9) = f(-2) means x = -9 and x = -2 give the same y-value, so the axis of symmetry sits halfway between them: h = (-9 + (-2))/2 = -11/2 = -5.5.

Find a. The axis of ax² - 7x + c is x = -(-7)/(2a) = 7/(2a). Set 7/(2a) = -11/2. Then 2a = -14/11 and a = -7/11, about -0.636. Since -0.636 is greater than -3/4 = -0.75, statement I is false.

Complete the square to find k.

1. f(x) = -7/11 (x² + 11x) + c
2. Half of 11 is 11/2, and (11/2)² = 121/4. Add and subtract it inside: f(x) = -7/11 (x² + 11x + 121/4 - 121/4) + c.
3. The -121/4 leaves as (-7/11)(-121/4) = +77/4: f(x) = -7/11 (x + 11/2)² + 77/4 + c.

So k = 77/4 + c = 19.25 + c. Because c > 0, k > 19.25, which is greater than 16. Statement II is true, and the answer is B.

Check on Desmos. Pick any positive c and test it:

1  a = -7/11
2  c = 1
3  f(x) = ax^2 - 7x + c
4  f(-9) - f(-2)
5  f(-5.5)

Line 4 returns 0, which confirms the symmetry. Line 5 returns 20.25, which is 19.25 + 1. Change line 2 to c = 5 and line 5 returns 24.25. A "must be true" statement has to hold for every allowed c, so the algebra decides it. Desmos only confirms the pattern.

Problem 4: a maximum in a word problem

A quadratic function z gives the estimated number of hours of daylight in a certain city t months after March 1. The maximum is 13.91 hours, 3.5 months after March 1. Six months after March 1, the estimate is 12.66 hours. What is the estimated number of hours of daylight on March 1?

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Translate. The maximum is the vertex, so the vertex is (3.5, 13.91) and z(t) = a(t - 3.5)² + 13.91. March 1 is t = 0, so the question asks for z(0).

On Desmos, start on a blank graph so the a from Problem 3 doesn't carry over. Line 1 is a table:

2  y_1 ~ a(x_1 - 3.5)^2 + 13.91
3  z(x) = a(x - 3.5)^2 + 13.91
4  z(0)

Line 2 fits a = -0.2. Line 4 returns 11.46 hours.

Check by hand. 12.66 = a(6 - 3.5)² + 13.91, so 6.25a = -1.25 and a = -0.2. Then z(0) = -0.2(12.25) + 13.91 = -2.45 + 13.91 = 11.46. Expanded, z(t) = -0.2t² + 1.4t + 11.46, and the constant term is the value at t = 0. For more problems that fit a model to a table, see SAT regression practice problems on Desmos.

Your turn: three SAT-style practice problems (you do)

We wrote these three for this page. Solve each one, then check it on Desmos. The answers are at the end of this section.

Practice 1. f(x) = ax² + 6x + c, where a and c are constants. If f(1) = f(-7) and the graph of y = f(x) passes through (0, 3), what is the vertex (h, k)?

Practice 2. A model gives the temperature T(d), in degrees Fahrenheit, d days after January 1: T(d) = a(d - 196)² + 92. The model gives 32°F at d = 0. What does the model give for T(365), to the nearest hundredth?

Practice 3. Write y = 3x² + 12x + 7 in vertex form. What is the vertex?

Answers

1. (-3, -6). f(1) = f(-7) puts the axis halfway between 1 and -7, so h = -3. The axis of ax² + 6x + c is x = -6/(2a), so -6/(2a) = -3 and a = 1. The point (0, 3) gives c = 3. Complete the square: x² + 6x + 3 = (x + 3)² - 9 + 3 = (x + 3)² - 6, so k = -6. On Desmos, graph y = x^2 + 6x + 3 and click the lowest gray dot.
2. About 47.39°F. T(0) = 32 gives a(196)² + 92 = 32, so a = -60/38416 = -15/9604, about -0.00156. Then T(365) = (-15/9604)(169)² + 92, about 47.39. On Desmos, type a = -15/9604, then T(x) = a(x - 196)^2 + 92, then T(365). Notice that T(365) is not 32, even though d = 365 is about a year after d = 0. A parabola doesn't repeat each year, so use a model like this only for the days it was built to cover.
3. y = 3(x + 2)² - 5, vertex (-2, -5). Factor out 3: 3(x² + 4x) + 7. Half of 4 is 2, and 2² = 4: 3(x² + 4x + 4) - 12 + 7 = 3(x + 2)² - 5. Graph both forms on Desmos and you see one parabola.

When algebra is faster

College Board's calculator policy says some Math questions are better done without a calculator. Vertex form answer choices are one case. In Problem 2 you can read the vertex off each choice in seconds, cross out D, and let one substitution pick B. Save Desmos for checking a conversion or for a table like Problem 4.

Mistakes to avoid

• Keeping the sign inside the parentheses. A vertex at (-2, 5) gives (x + 2)², not (x - 2)². Choice D in Problem 2 is built on this mistake, and it even passes through the given point.
• Forgetting the a when you complete the square. In 2(x² - 6x + 9 - 9) + 13, the -9 leaves as -18. If you write -9, you get 2(x - 3)² + 4, and Desmos draws a second parabola above the first.
• Mixing up the vertex and the start. In Problem 4, 13.91 is the maximum at t = 3.5. March 1 is t = 0. Write down what t = 0 means before you type anything.

Practice next

Study vertex form, standard form and factored form in the free lesson notes Quadratic Functions: Notes and Practice. For more SAT-style math sets, open the TTA SAT question bank.

For zeros, intercepts and graph questions, see SAT quadratics on Desmos. When a question asks for the sum of the solutions, use the sum of solutions formula. For the calculator itself, read our College Board Desmos guide.

Checked on September 26, 2026 against College Board's calculator policy and Desmos's help center articles FAQs and Regressions. The Test Advantage is independent of College Board and Desmos.