Unit 2, lesson 18

Quadratic Functions and Vertex Form

How to find the vertex, axis and zeros of a quadratic function by completing the square, factoring or the quadratic formula, and how the discriminant and the setting of a problem decide which answers count.

  • Theory part 35
  • Practice part 36
  • Quadratics, Circles, and Roots

Toss a ball and its height traces a curve that rises, turns over and falls. Two questions come up right away. How high does it go, and when does it land? The first asks for the vertex of a parabola and the second for its zeros, and both answers come from rewriting one quadratic expression in a more useful form.

Key terms

Quadratic function

Definition A quadratic function is a function of the form

f(x)=ax2+bx+c,f(x) = ax^2 + bx + c,

where aa, bb and cc are real constants and a≠0a \ne 0. Its domain is all real numbers unless a smaller domain is specified, and its graph is a curve called a parabola. The parabola opens upward when a>0a > 0 and downward when a<0a < 0.

In plain words The highest power of xx is 22. The sign of the leading coefficient aa decides whether the graph is a cup or a cap.

Example For f(x)=2x2−8x+6f(x) = 2x^2 - 8x + 6 the coefficients are a=2a = 2, b=−8b = -8 and c=6c = 6. Since a>0a > 0, the graph opens upward. A few values already hint at a lowest point and a symmetry.

xx0011223344
f(x)f(x)6600−2-20066

Standard and vertex forms

Definition The standard form of a quadratic function is f(x)=ax2+bx+cf(x) = ax^2 + bx + c with a≠0a \ne 0. The vertex form is

f(x)=a(x−h)2+k,a≠0.f(x) = a(x - h)^2 + k, \qquad a \ne 0.

Every quadratic function can be written in both forms with the same aa, as completing the square below shows, and the numbers hh and kk are

h=−b2a,k=f(h)=c−b24a.h = -\frac{b}{2a}, \qquad k = f(h) = c - \frac{b^2}{4a}.

In vertex form the graph is the graph of y=x2y = x^2 scaled vertically by the factor ∣a∣\lvert a \rvert, reflected across the xx-axis when a<0a < 0, and then shifted hh units to the right and kk units up. Some books call the vertex form the standard form and call ax2+bx+cax^2 + bx + c the general form, so check which convention a source uses.

In plain words Standard form shows the yy-intercept cc at a glance. Vertex form shows exactly where the graph turns.

Example Expanding a vertex form gives back the standard form.

2(x−2)2−2=2(x2−4x+4)−2=2x2−8x+6.\begin{aligned} 2(x - 2)^2 - 2 &= 2(x^2 - 4x + 4) - 2 \\[0.4em] &= 2x^2 - 8x + 6. \end{aligned}

The formulas agree, since

h=−−82⋅2=2,k=6−648=−2.\begin{gathered} h = -\frac{-8}{2 \cdot 2} = 2, \\[0.8em] k = 6 - \frac{64}{8} = -2. \end{gathered}

Vertex

Definition The vertex of the parabola y=a(x−h)2+ky = a(x - h)^2 + k is the point (h,k)(h, k). Because (x−h)2≥0(x - h)^2 \ge 0, with equality only at x=hx = h, the number kk is the minimum value of ff over all real xx when a>0a > 0 and the maximum value when a<0a < 0, and ff reaches it only at x=hx = h. On a smaller domain that leaves out hh, the value kk is never reached.

In plain words The vertex is the turning point, the bottom of a cup or the top of a cap.

Example For f(x)=2(x−2)2−2f(x) = 2(x - 2)^2 - 2 the vertex is (2,−2)(2, -2). Any other input adds the positive amount 2(x−2)22(x - 2)^2 to −2-2, so −2-2 is the minimum value. At x=5x = 5, for instance, the value is 2⋅9−2=162 \cdot 9 - 2 = 16.

Axis of symmetry

Definition The axis of symmetry of the parabola y=a(x−h)2+ky = a(x - h)^2 + k is the vertical line x=hx = h. The graph is symmetric about this line, which means

f(h+d)=f(h−d)for every real d,\begin{gathered} f(h + d) = f(h - d) \\[0.4em] \text{for every real } d, \end{gathered}

because both sides equal ad2+kad^2 + k.

In plain words Fold the graph along the vertical line through the vertex and the two halves match.

Example For f(x)=2x2−8x+6f(x) = 2x^2 - 8x + 6 the axis is x=2x = 2. The table above shows f(1)=f(3)=0f(1) = f(3) = 0 and f(0)=f(4)=6f(0) = f(4) = 6, pairs of inputs at equal distances from 22. Whenever two points of a parabola have the same height, the axis lies exactly halfway between them.

Completing the square

Definition For every real number bb,

x2+bx=(x+b2)2−b24.x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}.

To put f(x)=ax2+bx+cf(x) = ax^2 + bx + c in vertex form, factor aa out of the two terms with xx, complete the square inside the parentheses with this identity, and then collect the constants.

In plain words For positive numbers, picture x2+bxx^2 + bx as an xx by xx square with two strips, each b2\frac{b}{2} by xx, along two of its sides. The small corner square of side b2\frac{b}{2} is missing, and adding its area b24\frac{b^2}{4} completes a square of side x+b2x + \frac{b}{2}.

Example Take g(x)=−x2+6x−5g(x) = -x^2 + 6x - 5 and factor −1-1 out of the first two terms.

g(x)=−(x2−6x)−5=−[(x−3)2−9]−5=−(x−3)2+4.\begin{aligned} g(x) &= -(x^2 - 6x) - 5 \\[0.4em] &= -\left[(x - 3)^2 - 9\right] - 5 \\[0.4em] &= -(x - 3)^2 + 4. \end{aligned}

So the vertex is (3,4)(3, 4), the axis is x=3x = 3, and because a=−1<0a = -1 < 0 the maximum value of gg is 44.

Real zeros and factoring

Definition A real zero of a function ff is a real number rr with f(r)=0f(r) = 0. The real zeros are the first coordinates of the xx-intercepts (r,0)(r, 0) of the graph. Factoring a quadratic means writing it as

ax2+bx+c=a(x−r)(x−s)ax^2 + bx + c = a(x - r)(x - s)

with real numbers rr and ss, which is possible exactly when the quadratic has at least one real zero. A product of real numbers is zero only when one of the factors is zero, so the real zeros are then exactly rr and ss, a single number when r=sr = s.

In plain words Write the quadratic as a product. Each factor that can be zero marks a place where the graph meets the xx-axis.

Example Factor out the leading coefficient and then factor what is left.

2x2−8x+6=2(x2−4x+3)=2(x−1)(x−3).\begin{aligned} 2x^2 - 8x + 6 &= 2(x^2 - 4x + 3) \\[0.4em] &= 2(x - 1)(x - 3). \end{aligned}

The real zeros are 11 and 33, and the graph crosses the xx-axis at (1,0)(1, 0) and (3,0)(3, 0). The axis x=2x = 2 sits at their average, as the symmetry predicts.

Quadratic formula

Definition Let aa, bb and cc be real numbers with a≠0a \ne 0. If b2−4ac≥0b^2 - 4ac \ge 0, the real solutions of ax2+bx+c=0ax^2 + bx + c = 0 are

x=−b±b2−4ac2a,x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},

and if b2−4ac<0b^2 - 4ac < 0 the equation has no real solutions. The formula comes from completing the square in general, which turns the equation into

a(x+b2a)2=b2−4ac4a.a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}.

In plain words The formula finds the real solutions of any quadratic equation, including ones you cannot factor by eye. The plus and minus signs give two solutions at equal distances on either side of the axis of symmetry x=−b2ax = -\frac{b}{2a}, and they merge into one when the square root is 00.

Example The equation x2−4x−1=0x^2 - 4x - 1 = 0 has no factorization with integers. With a=1a = 1, b=−4b = -4 and c=−1c = -1,

x=4±16+42=2±5,x = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5},

which is about 4.2364.236 and −0.236-0.236. Both lie 5\sqrt{5} away from the axis x=2x = 2.

Discriminant

Definition The discriminant of ax2+bx+cax^2 + bx + c with a≠0a \ne 0 is the number D=b2−4acD = b^2 - 4ac. It sorts quadratics into three cases.

  • When D>0D > 0 there are two distinct real zeros, and the graph crosses the xx-axis twice.
  • When D=0D = 0 there is exactly one real zero, x=−b2ax = -\frac{b}{2a}, and the vertex lies on the xx-axis.
  • When D<0D < 0 there are no real zeros, and the graph never meets the xx-axis. The equation then has two solutions that are not real numbers, which come up in the lesson on complex numbers.

In plain words The number under the square root tells you how many real zeros to expect before you solve anything.

Example These three quadratics share the axis x=2x = 2 and differ only in the constant term.

QuadraticDiscriminantReal zeros
x2−4x+3x^2 - 4x + 316−12=416 - 12 = 411 and 33
x2−4x+4x^2 - 4x + 416−16=016 - 16 = 022 only
x2−4x+5x^2 - 4x + 516−20=−416 - 20 = -4none

Raising the constant lifts the parabola, so the two crossings move together, merge and then disappear.

Modeling constraints

Definition When a quadratic function models a situation, the model holds only on a domain set by that situation, such as nonnegative times or positive lengths. A vertex or zero computed from the formula answers the question only if it lies in that domain, and every answer should be checked against the original conditions and units.

In plain words The algebra does not know what xx stands for. You bring that knowledge in yourself and discard whatever does not fit.

Example A rectangular pen is built against a long wall, using 40 meters of fence for the other three sides. If each of the two sides at right angles to the wall has length xx, the side opposite the wall has length 40−2x40 - 2x, and the area is

A(x)=x(40−2x)=−2x2+40x,0<x<20.\begin{gathered} A(x) = x(40 - 2x) = -2x^2 + 40x, \\[0.4em] 0 < x < 20. \end{gathered}

The vertex is at x=−402⋅(−2)=10x = -\frac{40}{2 \cdot (-2)} = 10, so the largest possible area is A(10)=200A(10) = 200 square meters. The zeros 00 and 2020 lie outside the domain, because they describe a pen with no area at all. Setting A(x)=150A(x) = 150 gives x2−20x+75=0x^2 - 20x + 75 = 0, so x=5x = 5 or x=15x = 15. Since the parabola opens downward, the area is at least 150 square meters exactly when 5≤x≤155 \le x \le 15.

Check yourself

  1. Write f(x)=3x2+12x+7f(x) = 3x^2 + 12x + 7 in vertex form and give its vertex, axis of symmetry and minimum value.
Show the answer

Factor 33 out of the first two terms and complete the square.

3x2+12x+7=3(x2+4x)+7=3[(x+2)2−4]+7=3(x+2)2−5.\begin{aligned} &3x^2 + 12x + 7 \\[0.4em] &\quad = 3(x^2 + 4x) + 7 \\[0.4em] &\quad = 3\left[(x + 2)^2 - 4\right] + 7 \\[0.4em] &\quad = 3(x + 2)^2 - 5. \end{aligned}

The vertex is (−2,−5)(-2, -5), the axis is x=−2x = -2, and since 3>03 > 0 the minimum value is −5-5.

  1. How many real zeros does 2x2−3x+52x^2 - 3x + 5 have?
Show the answer

None. The discriminant is

(−3)2−4⋅2⋅5=9−40=−31,(-3)^2 - 4 \cdot 2 \cdot 5 = 9 - 40 = -31,

which is negative. This fits the graph, an upward parabola whose vertex (34,318)\left(\frac{3}{4}, \frac{31}{8}\right) lies above the xx-axis.

  1. A ball is thrown so that its height after tt seconds is h(t)=−5t2+20t+25h(t) = -5t^2 + 20t + 25 meters. Find its greatest height and the time it hits the ground.
Show the answer

The vertex is at

t=−202⋅(−5)=2,t = -\frac{20}{2 \cdot (-5)} = 2,

and h(2)=−20+40+25=45h(2) = -20 + 40 + 25 = 45, so the greatest height is 45 meters. Factoring gives

h(t)=−5(t2−4t−5)=−5(t−5)(t+1).\begin{aligned} h(t) &= -5(t^2 - 4t - 5) \\[0.4em] &= -5(t - 5)(t + 1). \end{aligned}

The zero t=−1t = -1 comes before the throw, so it lies outside the model. The ball lands at t=5t = 5 seconds.

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