Unit 3, lesson 20

The Quadratic Function and Its Graph

How to use the standard, vertex and factored forms of a quadratic to find its vertex, axis, intercepts, opening direction and range, and sketch the parabola from those facts.

  • Theory part 39
  • Practice part 40
  • Powers and Roots

Water from a garden hose, a ball tossed to a friend and the cable of a suspension bridge all trace curves shaped much like a parabola. A quadratic function is the simplest formula that bends, and it is the first curve calculus uses to show what a minimum, a maximum and a tangent line look like. The goal here is to sketch one from facts you can justify with algebra, not from a handful of plotted points and a guess.

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 with a≠0a \ne 0. Its domain is all real numbers. If a=0a = 0 the function is linear, not quadratic.

In plain words A quadratic has an x2x^2 term that really is there, and possibly an xx term and a constant.

Example The function f(x)=x2−4x−5f(x) = x^2 - 4x - 5 is quadratic with a=1a = 1, b=−4b = -4 and c=−5c = -5. The function g(x)=0x2+3x+1=3x+1g(x) = 0x^2 + 3x + 1 = 3x + 1 is not, because its x2x^2 coefficient is zero.

Equivalent forms

Definition A quadratic function can be written in three common forms that agree for every real xx. The standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c shows the yy intercept cc. The vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k shows the vertex (h,k)(h, k). Completing the square always produces it, with

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

The factored form f(x)=a(x−r1)(x−r2)f(x) = a(x - r_1)(x - r_2) shows the zeros r1r_1 and r2r_2. A factored form with real r1r_1 and r2r_2 exists exactly when ff has at least one real zero, that is, when b2−4ac≥0b^2 - 4ac \ge 0. If b2−4ac=0b^2 - 4ac = 0 the two factors are equal, r1=r2r_1 = r_2, and if b2−4ac<0b^2 - 4ac < 0 there is no factored form with real numbers.

The leading coefficient aa is the same in all three forms.

In plain words These are the same function wearing different clothes. Pick the form that shows the feature you need right now.

Example For f(x)=x2−4x−5f(x) = x^2 - 4x - 5, completing the square and factoring give

x2−4x−5=(x−2)2−9=(x−5)(x+1).\begin{aligned} x^2 - 4x - 5 &= (x - 2)^2 - 9 \\[0.4em] &= (x - 5)(x + 1). \end{aligned}

Expanding (x−2)2−9(x - 2)^2 - 9 gives x2−4x+4−9x^2 - 4x + 4 - 9, and expanding (x−5)(x+1)(x - 5)(x + 1) gives x2−4x−5x^2 - 4x - 5, so all three forms match.

Parabola

Definition The graph of a quadratic function, the set of points (x,ax2+bx+c)(x, ax^2 + bx + c) for real xx, is called a parabola. Every such graph is obtained from the graph of y=x2y = x^2 by a vertical stretch by the factor ∣a∣\lvert a \rvert, a reflection across the xx axis when a<0a < 0, and a shift that moves the origin to the vertex (h,k)(h, k).

In plain words Every quadratic graph is the basic y=x2y = x^2 curve, made narrower or wider, possibly flipped, and slid into place.

Example The graph of y=2(x−1)2+3y = 2(x - 1)^2 + 3 is the curve y=x2y = x^2 stretched vertically by 22 and moved 11 unit right and 33 units up. Its lowest point is (1,3)(1, 3). A table of values such as x=0,1,2x = 0, 1, 2 giving y=5,3,5y = 5, 3, 5 is a good check on a sketch, but a table alone cannot prove where the lowest point is. The vertex form proves it.

Opening direction

Definition For every real xx the square satisfies (x−h)2≥0(x - h)^2 \ge 0, with equality exactly when x=hx = h. Therefore in f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k the parabola opens upward when a>0a > 0, so that f(x)≥kf(x) \ge k for every xx, and opens downward when a<0a < 0, so that f(x)≤kf(x) \le k for every xx.

In plain words The sign of the x2x^2 coefficient decides everything about direction. Positive opens up like a cup, negative opens down like a cap.

Example The function g(x)=−x2+2x+3g(x) = -x^2 + 2x + 3 has a=−1<0a = -1 < 0, so its graph opens downward. In vertex form

g(x)=−(x−1)2+4,g(x) = -(x - 1)^2 + 4,

and since −(x−1)2≤0-(x - 1)^2 \le 0, every output satisfies g(x)≤4g(x) \le 4.

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). When a>0a > 0 the value kk is the minimum of the function f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, and when a<0a < 0 it is the maximum. In both cases it is attained only at x=hx = h, because (x−h)2=0(x - h)^2 = 0 only there.

In plain words The vertex is the turning point, the bottom of an upward parabola or the top of a downward one.

Example From f(x)=(x−2)2−9f(x) = (x - 2)^2 - 9 the vertex is (2,−9)(2, -9). Since a=1>0a = 1 > 0, the minimum value of ff is −9-9, reached only at x=2x = 2. The formula h=−b2ah = -\frac{b}{2a} agrees, giving −−42=2-\frac{-4}{2} = 2.

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. For every real tt,

f(h+t)=at2+k=f(h−t),f(h + t) = a t^2 + k = f(h - t),

so points at equal distances to the left and right of the axis have equal heights.

In plain words Fold the graph along the vertical line through the vertex and the two halves land on top of each other.

Example For f(x)=(x−2)2−9f(x) = (x - 2)^2 - 9 the axis is x=2x = 2. The inputs 00 and 44 are both 22 units from the axis, and indeed f(0)=−5f(0) = -5 and f(4)=−5f(4) = -5. Knowing the yy intercept (0,−5)(0, -5) therefore gives the mirror point (4,−5)(4, -5) for free.

Intercepts

Definition The yy intercept of f(x)=ax2+bx+cf(x) = ax^2 + bx + c is the point (0,c)(0, c), and every quadratic has exactly one. The xx intercepts are the points (r,0)(r, 0) where rr is a real solution of ax2+bx+c=0ax^2 + bx + c = 0. With the discriminant Δ=b2−4ac\Delta = b^2 - 4ac, there are two xx intercepts when Δ>0\Delta > 0, one (the vertex) when Δ=0\Delta = 0, and none when Δ<0\Delta < 0.

In plain words The graph always crosses the yy axis once. It meets the xx axis only where the equation f(x)=0f(x) = 0 has real solutions, which can be at two points, at one point where it just touches, or never.

Example For f(x)=x2−4x−5=(x−5)(x+1)f(x) = x^2 - 4x - 5 = (x - 5)(x + 1) the yy intercept is (0,−5)(0, -5) and the xx intercepts are (−1,0)(-1, 0) and (5,0)(5, 0). They sit symmetrically about the axis x=2x = 2, each 33 units away. For p(x)=x2+2x+5p(x) = x^2 + 2x + 5 the discriminant is 4−20=−16<04 - 20 = -16 < 0, so its graph never meets the xx axis, even though its yy intercept (0,5)(0, 5) exists.

Range

Definition A quadratic function f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k on all real numbers has range

{[k,∞)if a>0,(−∞,k]if a<0.\begin{cases} [k, \infty) & \text{if } a > 0, \\ (-\infty, k] & \text{if } a < 0. \end{cases}

The bound comes from (x−h)2≥0(x - h)^2 \ge 0. Every value beyond the bound is attained, since for yy with (y−k)/a≥0(y - k)/a \ge 0 the input

x=h+y−kax = h + \sqrt{\frac{y - k}{a}}

satisfies f(x)=yf(x) = y.

In plain words An upward parabola reaches every height from its vertex up, and a downward one reaches every height from its vertex down.

Example Take the downward parabola g(x)=−(x−1)2+4=−(x−3)(x+1)g(x) = -(x - 1)^2 + 4 = -(x - 3)(x + 1). Its outputs are at most 44. For any y≤4y \le 4 the input x=1+4−yx = 1 + \sqrt{4 - y} gives g(x)=−(4−y)+4=yg(x) = -(4 - y) + 4 = y, so the range is (−∞,4](-\infty, 4]. The full sketch now follows from justified features, the vertex (1,4)(1, 4) at the top, the axis x=1x = 1, the yy intercept (0,3)(0, 3) with its mirror point (2,3)(2, 3), and the xx intercepts (−1,0)(-1, 0) and (3,0)(3, 0).

Check yourself

  1. Write f(x)=2x2+8x+6f(x) = 2x^2 + 8x + 6 in vertex form and factored form. Give the vertex, opening direction, intercepts and range.
Show the answer

Completing the square and factoring give

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

The vertex is (−2,−2)(-2, -2) and the parabola opens upward because a=2>0a = 2 > 0. The yy intercept is (0,6)(0, 6), the xx intercepts are (−3,0)(-3, 0) and (−1,0)(-1, 0), and the range is [−2,∞)[-2, \infty).

  1. Does the graph of h(x)=−x2+6x−10h(x) = -x^2 + 6x - 10 cross the xx axis? Find its range.
Show the answer

In vertex form h(x)=−(x−3)2−1h(x) = -(x - 3)^2 - 1. The parabola opens downward with maximum value −1-1 at x=3x = 3, so every output is negative and the graph never meets the xx axis. The discriminant agrees, since 36−40=−4<036 - 40 = -4 < 0. The range is (−∞,−1](-\infty, -1].

  1. A parabola has vertex (1,−2)(1, -2) and passes through (3,6)(3, 6). Find its quadratic function.
Show the answer

Write f(x)=a(x−1)2−2f(x) = a(x - 1)^2 - 2. The point (3,6)(3, 6) gives 4a−2=64a - 2 = 6, so a=2a = 2 and f(x)=2(x−1)2−2f(x) = 2(x - 1)^2 - 2, which expands to 2x2−4x2x^2 - 4x.

How this lesson connects