How to Find a Quadratic's Minimum on Desmos

To find a quadratic's minimum on Desmos, graph it and click the lowest point of the parabola; its y-value is the minimum and its x-value is where it happens. A maximum works the same way on a parabola that opens down. SAT questions ask for one of those two numbers, so the skill that decides your answer is reading which one the question wants.

You can try every graph below in the College Board version of Desmos on our Desmos on the SAT page. This guide gives you the rule, three worked examples, four practice problems with answers and the by-hand method for each one.

The rule: the vertex holds both answers

The vertex is the turning point of a parabola. If the x² term has a positive coefficient, the parabola opens up and the vertex is its lowest point, the minimum. If the coefficient is negative, the parabola opens down and the vertex is the highest point, the maximum.

The vertex gives you two numbers, and SAT questions ask for them in different words:

• "What is the minimum value of f(x)?" wants the y-coordinate of the vertex.
• "For what value of x does f(x) reach its minimum?" wants the x-coordinate.

You can find the x-coordinate by hand from any form of the equation:

Then substitute that x-value into the function to get the minimum or maximum value. If your question comes in vertex form, our guide to SAT vertex form on Desmos covers reading h and k in more depth.

What Desmos shows you

The Desmos Help Center says that when you click anywhere on a line or curve, gray dots appear at intercepts, intersection points, maximums and minimums. Click a gray dot to display its coordinates, and click it again to hide them. College Board allows a calculator on the entire Math section, and Bluebook has a Desmos calculator built in with a graphing option, so you can use this on any quadratic question.

Worked example 1 (I do): minimum from factored form

SAT-style question. The function f is defined by f(x) = (x - 10)(x + 13). For what value of x does f(x) reach its minimum?

A) -130
B) -13
C) -23/2
D) -3/2

Step 1. Decide what you are looking for. The question says "for what value of x", so I need the x-coordinate of the vertex. Multiplying out gives x² + 3x - 130, and the positive x² term tells me the parabola opens up. It has a minimum.

Step 2. Graph it. I type the function on the first line exactly as it is written. There is no need to expand it.

1  y = (x - 10)(x + 13)

Step 3. Bring the vertex into view. The bottom of this parabola sits far below the starting view, so I click Zoom Out until I can see the bottom of the U. Desmos also lets you set the axes yourself in Graph Settings.

Step 4. Click the curve, then the lowest gray dot. The label reads the vertex: x = -1.5 and y = -132.25. The question wants x, and -1.5 = -3/2. The answer is D.

By-hand check. The zeros are x = 10 and x = -13. The vertex sits halfway between them: (10 + (-13))/2 = -3/2. Substitute to confirm the value: (-3/2 - 10)(-3/2 + 13) = (-23/2)(23/2) = -529/4 = -132.25.

Why the other choices are traps. -130 is f(0), the y-intercept. -13 is one of the zeros. -23/2 comes from averaging -10 and -13, which uses the numbers in the parentheses with the wrong sign. If the question had asked for the minimum value, the answer would be -132.25.

Worked example 2 (we do): maximum from standard form

SAT-style question. The function g is defined by g(x) = -2x² + 12x - 13. What is the maximum value of g(x)?

Step 1. The coefficient of x² is -2, which is negative. The parabola opens down, so the vertex is a maximum.

Step 2. The question asks for the maximum value, so you need the y-coordinate.

Step 3. Find the vertex x-value: x = -b/(2a) = -12/(2 × -2) = 3.

Step 4 (your step). Substitute x = 3 into g(x) and simplify.

Check your answer: g(3) = -2(9) + 12(3) - 13 = -18 + 36 - 13 = 5. The maximum value is 5, and it happens at x = 3.

To check on Desmos, type the function and click the top of the curve:

1  y = -2x^2 + 12x - 13

The gray dot at the top reads (3, 5). Type the exponent with ^ so Desmos reads x² correctly.

Want more questions like this one? Work through the Quadratic Functions lesson notes.

Worked example 3: maximum height in a word problem

SAT-style question. A ball is thrown upward. Its height, in feet, t seconds after it is thrown is h(t) = -16t² + 48t + 6. What is the maximum height of the ball, in feet?

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Desmos graphs in x and y, so type t as x:

1  y = -16x^2 + 48x + 6

Zoom out until the top of the curve shows, click the curve, and select the highest gray dot. It reads (1.5, 42). The ball reaches its highest point 1.5 seconds after the throw, and that height is 42 feet.

By hand: t = -48/(2 × -16) = 1.5, and h(1.5) = -16(2.25) + 72 + 6 = -36 + 72 + 6 = 42. Watch the wording again. "How many seconds after it is thrown does the ball reach its maximum height?" has the answer 1.5, not 42.

Your turn (you do): 4 practice problems

1. f(x) = (x + 4)(x - 8). What is the minimum value of f(x)?
2. p(x) = -x² + 10x + 11. What is the maximum value of p(x)?
3. The graph of y = 2x² - 12x + k, where k is a constant, has a minimum value of 5. What is the value of k?
4. A parabola opens downward and crosses the x-axis at (-7, 0) and (3, 0). For what value of x does the function reach its maximum?

Answers

1. -36. The zeros are -4 and 8, so the vertex is at x = 2, and f(2) = (6)(-6) = -36. On Desmos, the lowest gray dot reads (2, -36).
2. 36. x = -10/(2 × -1) = 5, and p(5) = -25 + 50 + 11 = 36. Factoring gives -(x - 11)(x + 1), and the average of 11 and -1 is also 5.
3. 23. The vertex is at x = -(-12)/(2 × 2) = 3, and y(3) = 18 - 36 + k = k - 18. Set k - 18 = 5 to get k = 23. Graph y = 2x^2 - 12x + 23 and the lowest gray dot reads (3, 5).
4. -2. The maximum sits halfway between the x-intercepts: (-7 + 3)/2 = -2. You don't need the equation for this one.

When algebra is faster

College Board's calculator guidance says some Math questions are better done without a calculator, even though one is allowed. Problem 4 is one of them: two x-intercepts and one average give the answer. When the quadratic is already in factored or vertex form, the by-hand route is often shorter than zooming. Use Desmos to confirm, or when the equation is messy.

Mistakes to avoid

• Answering with the wrong coordinate. "Minimum value" is the y-coordinate. "For what value of x" is the x-coordinate. Circle the phrase before you click anything.
• Flipping the signs of the zeros. (x - 10)(x + 13) has zeros 10 and -13. Reading them as -10 and 13 puts the vertex at x = 3/2 instead of -3/2.
• Reading a dot that is not the vertex. Gray dots also mark the intercepts. The vertex is the dot at the turning point. If you cannot see a turning point, zoom out before you read anything.

Practice next

Start with the Quadratic Functions lesson notes. For solving quadratics rather than graphing them, try the Quadratic Equations lesson notes. For more questions, browse the SAT question bank.

Related guides: for more quadratic questions on the calculator, read Digital SAT Desmos quadratics. If a question asks for the sum of the roots instead of the vertex, see the sum of solutions formula.

Checked on September 26, 2026 against College Board's SAT Suite calculator policy, the College Board version of the Desmos graphing calculator and the Desmos Help Center FAQs on important points and the viewing window. Every answer on this page was recomputed before publishing. The Test Advantage is independent of College Board and Desmos.