Unit 9 · Lesson 4 of 5
Graphing parabolas and max/min word problems
You'll be able to: find a parabola's vertex, axis, and intercepts, sketch it, and use the vertex to answer "highest" or "largest" word problems.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Both come back today as intercepts.
Solve with the quadratic formula: \(x^2 + 2x - 2 = 0\).
\(a = 1\), \(b = 2\), \(c = -2\). Discriminant: \(4 - 4(1)(-2) = 4 + 8 = 12\), and \(\sqrt{12} = 2\sqrt{3}\).
\(x = \dfrac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}\).
Find the y-intercept of the line \(3x - 2y = 12\). Give it as a point.
Set \(x = 0\): \(-2y = 12\), so \(y = -6\).
Answer: \((0, -6)\).
Step 2 of 5The idea
The graph of \(y = ax^2 + bx + c\) is a parabola, a U shape. Its turning point is the vertex. Find the vertex first, and the rest of the graph falls into place.
Direction. \(a \gt 0\): opens up, and the vertex is the lowest point (a minimum). \(a \lt 0\): opens down, and the vertex is the highest point (a maximum).
Vertex. Its x-value is \(x = -\dfrac{b}{2a}\). Put that x into the equation to get its y-value.
Axis of symmetry. The vertical line \(x = -\dfrac{b}{2a}\) through the vertex. The two halves are mirror images.
Intercepts. y-intercept: \((0, c)\). x-intercepts: solve \(ax^2 + bx + c = 0\). There can be two, one, or none.
Vertex form. \(y = a(x - h)^2 + k\) has its vertex at \((h, k)\). For example, \(y = -2(x - 1)^2 + 8\) has vertex \((1, 8)\) and opens down.
The line through the vertex is the axis of symmetry, and \(a(x - h)^2 + k\) is called vertex form.
Word problems that ask for the "maximum height", "largest area", or "most revenue" are vertex questions. The vertex's x-value says when or which input; its y-value is the biggest (or smallest) amount.
Why this shows up on the placement test: expect at least one "find the vertex" or "what is the maximum height" question; it's a favorite.
Step 3 of 5Worked examples
Example 1: Graph \(y = x^2 - 4x + 3\)
- \(a = 1\) is positive, so it opens up. The vertex will be the lowest point.
- Vertex x-value: \(x = -\dfrac{-4}{2(1)} = 2\). Its y-value: \(2^2 - 4(2) + 3 = 4 - 8 + 3 = -1\). Vertex: \((2, -1)\).
- Axis of symmetry: \(x = 2\).
- y-intercept: \(c = 3\), so \((0, 3)\). Its mirror image across \(x = 2\) is \((4, 3)\).
- x-intercepts: \(x^2 - 4x + 3 = (x - 1)(x - 3) = 0\), so \((1, 0)\) and \((3, 0)\). Each is 1 unit from the axis, as the mirror rule says.
Example 2: A maximum height word problem
A ball is thrown upward. Its height in feet after \(t\) seconds is \(h = -16t^2 + 64t + 5\). When is it highest, and how high does it get?
- \(a = -16\) is negative, so the parabola opens down. The vertex is the highest point.
- Vertex t-value: \(t = -\dfrac{64}{2(-16)} = -\dfrac{64}{-32} = 2\) seconds.
- Height then: \(h = -16(2)^2 + 64(2) + 5 = -64 + 128 + 5 = 69\) feet.
- Answer: the ball is highest after 2 seconds, at 69 feet. The first coordinate answers "when", the second answers "how high".
Step 4 of 5Practice
Type your answer and press Check. Write points like (2, -3). If you miss twice, open the worked solution. Then mark how it went.
What is the vertex of \(y = 2(x + 1)^2 - 5\)?
This is vertex form \(a(x - h)^2 + k\). Since \(x + 1 = x - (-1)\), \(h = -1\). And \(k = -5\).
Answer: \((-1, -5)\).
Which parabola opens down and has its vertex at \((2, -3)\)?
- \(y = -(x + 2)^2 - 3\)
- \(y = (x - 2)^2 - 3\)
- \(y = -(x - 2)^2 - 3\)
- \(y = -(x - 2)^2 + 3\)
Opens down needs a negative \(a\). Vertex \((2, -3)\) needs \((x - 2)\) inside and \(-3\) outside.
(a) has vertex \((-2, -3)\). (b) opens up. (d) has vertex \((2, 3)\). Answer: (c).
Find the vertex of \(y = -3x^2 + 6x + 1\).
\(a = -3\), \(b = 6\). x-value: \(x = -\dfrac{6}{2(-3)} = -\dfrac{6}{-6} = 1\).
y-value: \(-3(1)^2 + 6(1) + 1 = -3 + 6 + 1 = 4\).
Answer: \((1, 4)\).
Find the axis of symmetry of \(y = 2x^2 + 12x - 1\).
\(a = 2\), \(b = 12\). \(x = -\dfrac{12}{2(2)} = -\dfrac{12}{4} = -3\).
Answer: the line \(x = -3\).
At which x-values does the graph of \(y = x^2 + 2x - 8\) cross the x-axis?
Set \(y = 0\): \(x^2 + 2x - 8 = 0\).
Factor: \((x + 4)(x - 2) = 0\), so \(x = -4\) or \(x = 2\).
On paper, sketch \(y = -x^2 + 2x + 3\). Label the vertex, the y-intercept, and the x-intercepts. Then show the answer to compare.
Opens down. Vertex \((1, 4)\), y-intercept \((0, 3)\), x-intercepts \((-1, 0)\) and \((3, 0)\).
\(a = -1\), so it opens down. Vertex x-value: \(x = -\dfrac{2}{2(-1)} = 1\); y-value: \(-1 + 2 + 3 = 4\).
y-intercept: \(c = 3\), so \((0, 3)\).
x-intercepts: \(-x^2 + 2x + 3 = 0\). Multiply by \(-1\): \(x^2 - 2x - 3 = 0\), so \((x - 3)(x + 1) = 0\): \(x = 3\) or \(x = -1\).
A ball's height in feet after \(t\) seconds is \(h = -16t^2 + 48t + 4\). What is its maximum height, in feet?
Opens down (\(a = -16\)), so the vertex is the maximum. \(t = -\dfrac{48}{2(-16)} = \dfrac{48}{32} = 1.5\) seconds.
\(h = -16(1.5)^2 + 48(1.5) + 4 = -16(2.25) + 72 + 4 = -36 + 76 = 40\).
Answer: \(40\) feet.
You have 80 feet of fence for a rectangular pen. If one side is \(x\) feet long, the area is \(A = x(40 - x)\) square feet. What is the largest possible area?
Multiply out: \(A = -x^2 + 40x\). Here \(a = -1\), \(b = 40\), so it opens down.
Vertex: \(x = -\dfrac{40}{2(-1)} = 20\) feet.
\(A = 20(40 - 20) = 20 \cdot 20 = 400\). Answer: \(400\) square feet (a 20 by 20 square).
Step 5 of 5Wrap-up
You can now read a parabola: the sign of \(a\) gives the direction, \(x = -\frac{b}{2a}\) gives the vertex and axis, \(c\) gives the y-intercept, and solving \(= 0\) gives the x-intercepts. In word problems, the vertex is the maximum or minimum.
Read or watch next: OpenStax Intermediate Algebra 2e, 9.6 Graph Quadratic Functions Using Properties (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Vertex form (short videos). Quick rules: Quadratics cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why does \(x = -\frac{b}{2a}\) give the vertex?" It knows this course and can explain things another way.