SAT Quadratic Vertex Word Problems: Maximum or Minimum?

Find a quadratic vertex, interpret both coordinates with units, and check time ranges or whole-number inputs. Includes original worked examples and practice.

Find the vertex, check whether its input is allowed, and answer with the coordinate the question requests. A downward-opening parabola has a maximum at its vertex; an upward-opening parabola has a minimum. The input tells you when or where that value occurs, while the output tells you the value itself.

College Board’s digital SAT framework includes creating and using quadratic functions and interpreting their features in context. The problems below are original Kibo practice. Their fictional models are supplied for the exercises; they are not official SAT items or measured scientific data.

Read the two coordinates separately

In vertex form, f(x)=a(x−h)2+kf(x)=a(x-h)^2+k, with a≠0a\neq0, the vertex is (h,k)(h,k). The squared term becomes zero at x=hx=h, leaving f(h)=kf(h)=k. If a<0a<0, every other input subtracts from kk; if a>0a>0, every other input adds to it.

What the question asks you to report
Requested quantityAnswer to find
When or where the maximum or minimum occursThe input hh, with input units
The greatest or least valueThe output kk, with output units
The vertex coordinatesThe ordered pair (h,k)(h,k)

These vertex answers apply when that input belongs to the stated domain, meaning the allowed inputs. OpenStax’s quadratic-functions lesson explains vertex form, the direction a parabola opens, and how a situation can restrict the domain.

Worked maximum: how deep does a sensor go?

An underwater sensor’s depth D(t)D(t), in meters below the water surface, is modeled by the following function. Time tt is measured in minutes, and the model applies for 0≤t≤80\le t\le8. What is the greatest depth, and when does it occur?

D(t)=−2(t−4)2+35D(t)=-2(t-4)^2+35

The negative coefficient −2-2 makes the parabola open downward. At t=4t=4, the squared term is zero, so D(4)=35D(4)=35. This time is within the allowed interval. The sensor’s greatest modeled depth is 3535 meters, reached 44 minutes after the start.

The vertex (4,35)(4,35) pairs a time with a depth. If the question asks only for the greatest depth, report 3535 meters. If it asks how long after the start that depth occurs, report 44 minutes.

Checking the sensor model at selected times
Time in minutesDepth in meters
00−2(0−4)2+35=3-2(0-4)^2+35=3
33−2(3−4)2+35=33-2(3-4)^2+35=33
44−2(4−4)2+35=35-2(4-4)^2+35=35
55−2(5−4)2+35=33-2(5-4)^2+35=33
88−2(8−4)2+35=3-2(8-4)^2+35=3

The matching outputs at 33 and 55 minutes reflect symmetry around t=4t=4. The table checks particular values; the nonnegative square proves that no allowed time can produce a depth greater than 3535 meters.

What changes if the question covers only the first three minutes?

Keep the same formula, but restrict the question to 0≤t≤30\le t\le3. Now t=4t=4 is outside the interval. Over the allowed times, moving toward 33 makes (t−4)2(t-4)^2 smaller and the depth larger. The greatest depth during this shorter interval is D(3)=33D(3)=33 meters, reached at 33 minutes.

For a quadratic on a closed interval L≤x≤UL\le x\le U, compare the outputs at both included endpoints and at the vertex if its input is inside the interval. Choose the greatest or least of those outputs as requested. An excluded endpoint cannot supply an attained maximum or minimum.

Worked minimum: start with an expanded quadratic

In a fictional cooling-and-warming model, a sample’s temperature T(t)T(t), in degrees Celsius, is given below for 0≤t≤50\le t\le5, with time in minutes. Find the minimum temperature and the time when it occurs.

T(t)=2t2−12t+29T(t)=2t^2-12t+29

For f(x)=ax2+bx+cf(x)=ax^2+bx+c, the vertex input is h=−b2ah=-\frac{b}{2a}. Here a=2a=2 and b=−12b=-12, so:

h=−−122(2)=3h=-\frac{-12}{2(2)}=3

Evaluate the original function: T(3)=2(3)2−12(3)+29=11T(3)=2(3)^2-12(3)+29=11. The positive squared-term coefficient means the vertex is a minimum, and 33 is an allowed time. The minimum temperature is 1111 degrees Celsius after 33 minutes.

Completing the square gives an independent way to see it: T(t)=2(t−3)2+11T(t)=2(t-3)^2+11. The squared term can contribute nothing smaller than zero. Endpoint checks give T(0)=29T(0)=29 and T(5)=19T(5)=19, both above 1111.

Notice that the constant 2929 gives T(0)T(0), the initial temperature. It does not give the minimum. If signs or squaring a negative value are causing trouble, review our guide to negative signs and parentheses; those algebra rules also apply here.

Whole-number inputs can change the answer

A display plan uses nn rows of cards, with 13−n13-n cards in each row. The allowed row counts are the integers from 22 through 1010. How many cards can the display hold at most?

C(n)=n(13−n)=−n2+13nC(n)=n(13-n)=-n^2+13n

The continuous parabola has its vertex at n=−132(−1)=6.5n=-\frac{13}{2(-1)}=6.5, with output 42.2542.25. A display cannot have 6.56.5 whole rows. Check the allowed integers on either side: C(6)=6(7)=42C(6)=6(7)=42 and C(7)=7(6)=42C(7)=7(6)=42. Both arrangements hold the maximum of 4242 cards.

The form C(n)=−(n−6.5)2+42.25C(n)=-(n-6.5)^2+42.25 explains why: the closest allowed integers to 6.56.5 make the squared distance smallest. For a downward-opening quadratic over consecutive allowed integers, check the nearest allowed integers to the vertex. If the vertex lies beyond the allowed range, check the nearest allowed endpoint. Rounding the output alone does not identify a valid input.

Try two fresh problems

Before calculating, identify the allowed inputs and whether the question asks for an input or an output.

  1. For g(x)=3(x+2)2−7g(x)=3(x+2)^2-7, with all real inputs allowed, what is the minimum value, and at which input does it occur?
  2. A curved panel has height H(x)=−x2+10x+2H(x)=-x^2+10x+2 meters above a baseline, where xx is horizontal distance in meters. Only the piece with 0≤x≤40\le x\le4 is installed. What is the greatest height of the installed piece, and where does it occur?
Show answers and checks
  1. The minimum value is −7-7, at x=−2x=-2. The expression x+2x+2 equals zero when x=−2x=-2. Since 3(x+2)2≥03(x+2)^2\ge0, no input gives an output below −7-7. Substitution confirms g(−2)=−7g(-2)=-7.
  2. The vertex of the full parabola is at x=5x=5, with height 2727, but that input is outside the installed piece. In vertex form, H(x)=−(x−5)2+27H(x)=-(x-5)^2+27, so height increases as the allowed input moves toward 44. The endpoint values are H(0)=2H(0)=2 and H(4)=−16+40+2=26H(4)=-16+40+2=26. The greatest installed height is 2626 meters at x=4x=4 meters.

Check the meaning before submitting

  • Use the input that makes the squared expression zero. In (x+2)2(x+2)^2, that input is −2-2.
  • Label each coordinate with its quantity and units. Time, depth, temperature, and a count of cards answer different questions.
  • Check the domain before reporting the vertex. A time limit or a whole-number requirement can rule it out.
  • Use the sign of aa to decide whether the vertex gives a maximum or a minimum. Solving f(x)=0f(x)=0 finds zeros, which answers a different question.

Record whether a missed problem came from the vertex calculation, the domain, or the requested quantity in our practice error log. Rework that step, then check a different problem without its solution visible.

Try a free SAT practice question with Kibo.

Prepared with AI assistance. Original examples were checked with separate exact calculations; no human review is claimed. Sources checked October 29, 2026. SAT is a registered trademark of the College Board, which is not affiliated with or endorsing this guide.