Quadratic Equations: PSAT/NMSQT Practice Problems

Quadratic equations appear throughout the Passport to Advanced Math section of the PSAT/NMSQT, and one of the most heavily tested skills is recognizing that the same parabola can be written in three different but equivalent forms — standard form, factored form, and vertex form — and that each form makes a different piece of information easy to read directly off the equation. Standard form (y = ax² + bx + c) makes the y-intercept obvious. Factored form (y = a(x − p)(x − q)) makes the x-intercepts obvious. Vertex form (y = a(x − h)² + k) makes the vertex — the parabola's maximum or minimum point — obvious. The PSAT often asks you to convert between these forms specifically because a question wants a piece of information that isn't visible in the form the equation started in.

Converting to vertex form usually means completing the square, and converting to factored form usually means factoring the quadratic expression directly. A related and very common question type gives you a parabola's vertex form or factored form with an unknown coefficient a, along with one additional point the parabola passes through, and asks you to solve for a by substitution. Because a controls how narrow or wide (and whether the parabola opens upward or downward) the graph is, these questions test whether you can correctly plug a known ordered pair into the equation and solve algebraically for that one remaining unknown.

Below are 5 PSAT/NMSQT-style practice problems covering vertex form, factored form, and solving for an unknown coefficient using a known point on the parabola. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free PSAT/NMSQT Math QBank.

Questions 1 and 2 refer to the following information. The graph of quadratic function y = x² − 6x + 5 is shown below.

Converting to vertex form

Which of the following is an equivalent form of the equation of the graph shown above, from which the coordinates of vertex V can be identified as constants in the equation?

Full Explanation

The vertex is visible as constants only in vertex form, y = a(x − h)² + k, where (h, k) is the vertex. Complete the square on y = x² − 6x + 5 to get there.

Take half of the x-coefficient (−6) and square it: (−6/2)² = 9. Add and subtract 9 to keep the expression equivalent:

y = (x² − 6x + 9) − 9 + 5 = (x − 3)² − 4

This matches choice D, with vertex (3, −4) — consistent with the graph, which shows V below and between the x-intercepts. The correct answer is D.

Converting to factored form

Which of the following is an equivalent form of the equation of the graph shown above, that displays the x-intercepts of the parabola as constants?

Full Explanation

The x-intercepts are visible as constants only in factored form, y = a(x − p)(x − q), where p and q are the x-intercepts. Factor x² − 6x + 5 by finding two numbers that multiply to 5 and add to −6: those numbers are −1 and −5.

y = (x − 1)(x − 5)

This matches the graph's x-intercepts at x = 1 and x = 5. Be careful with the sign: the factors use minus the intercept value, so intercepts of 1 and 5 give (x − 1) and (x − 5), not choice B.

The correct answer is A.

Solving for a using a double root and the y-intercept

In the xy-plane above, the parabola y = a(x − h)² has one x-intercept at (4, 0). If the y-intercept of the parabola is 9, what is the value of a?

Full Explanation

Vertex form y = a(x − h)² (with no added constant) has only one x-intercept, which means that x-intercept is also the vertex. So the x-intercept (4, 0) tells us h = 4: y = a(x − 4)².

The y-intercept is the value of y when x = 0. Substitute x = 0 and set y equal to the given y-intercept, 9:

9 = a(0 − 4)² = 16a

a = 9/16. The correct answer is B.

Solving for a using vertex form and a known point

In the xy-plane, if the parabola with equation y = a(x + 2)² − 15 passes through (1, 3), what is the value of a?

Full Explanation

"Passes through (1, 3)" means x = 1 and y = 3 must satisfy the equation. Substitute both values in:

3 = a(1 + 2)² − 15

3 = a(3)² − 15 = 9a − 15

3 + 15 = 9a → 18 = 9a → a = 2. The correct answer is B.

Solving for a using factored form and the minimum value

The graph of the equation y = a(x − 1)(x + 5) is a parabola with vertex (h, k). If the minimum value of y is −12, what is the value of a?

Full Explanation

The factored form shows the x-intercepts directly: x = 1 and x = −5. A parabola's vertex always lies exactly halfway between its two x-intercepts (its axis of symmetry):

h = (1 + (−5))/2 = −2

Since the problem says y has a minimum value, k is that minimum: k = −12. Substitute the vertex's x-coordinate into the original equation and set y equal to −12 to solve for a:

−12 = a(−2 − 1)(−2 + 5) = a(−3)(3) = −9a

a = 12/9 = 4/3. (Note a is positive, which makes sense — a parabola only has a minimum, rather than a maximum, when it opens upward.) The correct answer is B.

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