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Nonlinear functions 1: quadratics and parabolas

SAT Math Β· Advanced Math Β· Week 6 of the 12-week plan
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Nonlinear functions 1: quadratics and parabolas

College Board skill: Nonlinear functions
GoalRead the vertex, axis of symmetry, zeros, y-intercept and maximum or minimum of a quadratic function from its standard, vertex or factored form, and use these ideas in real-life situations.
On the test

Advanced Math is about 35% of SAT Math (13–15 of 44 questions), and quadratic functions are one of its biggest topics. Expect questions about the vertex, the minimum or maximum value, the zeros, equivalent forms, and stories such as the height of a thrown ball.

Key words
parabola Β· the U-shaped graph of a quadratic function y = axΒ² + bx + cvertex Β· the turning point of the parabola: its lowest or highest pointaxis of symmetry Β· the vertical line through the vertex; the two halves are mirror imageszero Β· an x-value where the function equals 0 (an x-intercept of the graph)
Explanation

Three forms, three facts

The same quadratic function can be written in three ways, and each form shows something different at a glance. Choose the form that shows what the question asks for, or rewrite the function into it.

FormEquationWhat you see at once
Standardy = axΒ² + bx + cy-intercept c; vertex at x = βˆ’b/(2a)
Vertexy = a(x βˆ’ h)Β² + kvertex (h, k); axis x = h
Factoredy = a(x βˆ’ p)(x βˆ’ q)zeros p and q

Vertex and axis of symmetry

The axis of symmetry is the vertical line x = βˆ’b/(2a). The vertex lies on it. To get the y-coordinate of the vertex, put the x-value into the function. In the factored form the axis is exactly halfway between the zeros: x = (p + q)/2. For y = xΒ² βˆ’ 2x βˆ’ 3 = (x βˆ’ 3)(x + 1) the zeros are βˆ’1 and 3, the axis is x = 1, and the vertex is (1, βˆ’4).

xyβˆ’3βˆ’2βˆ’112345βˆ’6βˆ’5βˆ’4βˆ’3βˆ’2βˆ’1123450y = xΒ² βˆ’ 2x βˆ’ 3(βˆ’1, 0)(3, 0)vertex (1, βˆ’4)

Opens up or down: maximum or minimum

The sign of a decides the direction. If a > 0, the parabola opens upward and the vertex is the lowest point: the function has a minimum value k. If a < 0, it opens downward and the vertex is the highest point: the function has a maximum value k. The value k is the y-coordinate of the vertex, not the x-coordinate. Many questions ask β€œwhat is the maximum value”, so give the y-value.

Changing the form

From factored to standard: expand. From standard to vertex form: complete the square. Take f(x) = 2xΒ² βˆ’ 12x + 10. Pull out the 2 from the x-terms: 2(xΒ² βˆ’ 6x) + 10. Half of βˆ’6 is βˆ’3, and (βˆ’3)Β² = 9, so xΒ² βˆ’ 6x = (x βˆ’ 3)Β² βˆ’ 9. Then f(x) = 2((x βˆ’ 3)Β² βˆ’ 9) + 10 = 2(x βˆ’ 3)Β² βˆ’ 8. The vertex is (3, βˆ’8). Remember to multiply the βˆ’9 by the 2.

Quadratics in real life

The height of a thrown object is a quadratic in time, for example h(t) = βˆ’16tΒ² + 48t + 4 with h in feet and t in seconds. The vertex gives the highest point and the time it is reached. The positive zero gives the time the object hits the ground. The y-intercept h(0) is the starting height. A farmer’s rectangle with a fixed fence length also gives a quadratic: the maximum area is at the vertex.

t, secondsh, feet123410203040500h(t)highest point (1.5, 40)start (0, 4)
Worked examples
Example 1.
What is the minimum value of f(x) = xΒ² βˆ’ 6x + 5?
  1. The parabola opens upward (a = 1), so it has a minimum at the vertex.
  2. x-coordinate: βˆ’b/(2a) = 6/2 = 3.
  3. f(3) = 9 βˆ’ 18 + 5 = βˆ’4.
  4. The minimum value is βˆ’4 (it occurs at x = 3).
Trap: The answer is the y-value βˆ’4, not the x-value 3.
Example 2.
The function f(x) = 2(x + 1)(x βˆ’ 5) is graphed in the xy-plane. What is the vertex of the graph?
  1. (βˆ’2, βˆ’18)
  2. (2, βˆ’18)
  3. (2, 18)
  4. (3, βˆ’16)
  1. The zeros are x = βˆ’1 and x = 5.
  2. The axis of symmetry is halfway: x = (βˆ’1 + 5)/2 = 2.
  3. f(2) = 2(3)(βˆ’3) = βˆ’18.
  4. The vertex is (2, βˆ’18).
Trap: The point (3, βˆ’16) lies on the graph, but it is not the vertex.
Example 3.
A parabola has the vertex (3, 8) and passes through the point (5, 0). The equation is y = a(x βˆ’ 3)Β² + 8. What is the value of a?
xyβˆ’11234567βˆ’4βˆ’22468100parabolavertex (3, 8)(5, 0)
  1. Substitute the point (5, 0): 0 = a(5 βˆ’ 3)Β² + 8.
  2. 0 = 4a + 8, so a = βˆ’2.
  3. Since a < 0, the parabola opens downward, which matches a vertex that is higher than the point (5, 0).
Example 4.
A ball is thrown upward. Its height, in feet, after t seconds is h(t) = βˆ’16tΒ² + 48t + 4. What is the maximum height of the ball, in feet?
  1. a = βˆ’16 < 0, so the vertex is the highest point.
  2. t = βˆ’b/(2a) = βˆ’48/(βˆ’32) = 1.5 seconds.
  3. h(1.5) = βˆ’16(2.25) + 48(1.5) + 4 = βˆ’36 + 72 + 4 = 40.
Common traps
  • Giving the x-value of the vertex instead of the maximum or minimumThe maximum or minimum value is the y-coordinate k. The x-coordinate tells where it happens.
  • Wrong sign in vertex formy = a(x βˆ’ h)Β² + k has vertex (h, k). So y = (x + 2)Β² βˆ’ 5 has vertex (βˆ’2, βˆ’5), not (2, βˆ’5).
  • Forgetting to multiply when completing the square2(xΒ² βˆ’ 6x) = 2((x βˆ’ 3)Β² βˆ’ 9). The βˆ’9 must also be multiplied by 2.
  • Reading the zeros with the wrong signsy = (x βˆ’ 4)(x + 1) has zeros 4 and βˆ’1: set each factor to 0.
  • Using the factored form to read the y-interceptThe y-intercept is f(0). In y = 2(x + 1)(x βˆ’ 5) it is 2(1)(βˆ’5) = βˆ’10, not 2.
  • Mixing up time and height in a storyIn h(t), t is the horizontal axis. β€œWhen” asks for t; β€œhow high” asks for h.
The Desmos way

Type the function and Desmos draws the parabola. Click on the curve at the vertex or at an x-intercept and the coordinates appear, with the lowest or highest point and the zeros marked by grey dots. To test a vertex-form equation, type y = a(x βˆ’ h)Β² + k and accept the sliders for a, h and k; moving them shows how each one moves or flips the parabola. You can also type a table of values to see whether the points lie on a parabola.

  1. Type y = βˆ’16x^2 + 48x + 4
  2. Click the highest point: it is (1.5, 40)
  3. Click the right x-intercept to find when the object lands (about 3.08)
  4. Type y = a(x βˆ’ h)^2 + k and move the sliders to match a given graph

Use Desmos to find a vertex or zeros in one click, especially for ugly numbers. Know the vertex formula too, since it is quick when b and a are small integers.

Open Desmos β†—
Quick check
1
What is the vertex of y = (x + 1)Β² βˆ’ 6?
2
What are the zeros of y = (x βˆ’ 2)(x + 4)?
3
What is the axis of symmetry of y = xΒ² + 8x + 3?
4
Does y = βˆ’3xΒ² + 1 have a maximum or a minimum? What is its value?
Practice set: 10 SAT-style questionsEasy β†’ hard, with typed answers like the real test. Your score is saved in your cabinet.
Start practice β†’

More official practice

Bluebook practice tests β†—Official full-length adaptive tests from the College Board, free SAT question bank β†—Official questions you can filter by skill and difficulty Khan Academy β†—Free official SAT lessons and practice Desmos calculator β†—The same graphing calculator that is built into the test