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Quadratic Functions

Constructing Quadratic Functions

How to Determine Quadratic Function Equations

Constructing a quadratic function means determining the form of the equation f(x)=ax2+bx+cf(x) = ax^2 + bx + c based on given information. There are several different ways to construct a quadratic function, depending on the available information.

Forms of Quadratic Functions

Before we start constructing quadratic functions, let's understand the three forms of quadratic functions:

  1. Standard Form: f(x)=ax2+bx+cf(x) = ax^2 + bx + c
  2. Factored Form: f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q) where p and q are the roots of the equation
  3. Vertex Form: f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k where (h, k) is the vertex point

These three forms are interrelated and can be converted from one form to another.

Methods of Constructing Quadratic Functions

Three Points

The most common way to construct a quadratic function is by using three known points on the curve.

If we have three points (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3), we can substitute these values into the standard equation to get three linear equations with three variables a, b, and c.

y1=ax12+bx1+cy_1 = ax_1^2 + bx_1 + c
y2=ax22+bx2+cy_2 = ax_2^2 + bx_2 + c
y3=ax32+bx3+cy_3 = ax_3^2 + bx_3 + c

Example:

Find the quadratic function that passes through points K(-1, 0), L(0, -3), and M(1, -4).

Substitute the coordinate values into the standard equation.

K(−1,0):0=a(−1)2+b(−1)+cK(-1, 0): 0 = a(-1)^2 + b(-1) + c
L(0,−3):−3=a(0)2+b(0)+cL(0, -3): -3 = a(0)^2 + b(0) + c
M(1,−4):−4=a(1)2+b(1)+cM(1, -4): -4 = a(1)^2 + b(1) + c

Simplify these equations.

0=a−b+c0 = a - b + c
−3=c-3 = c
−4=a+b+c-4 = a + b + c

Substitute c=−3c = -3 into the first and third equations.

0=a−b−30 = a - b - 3
−4=a+b−3-4 = a + b - 3

Simplify and add the two equations to find the value of a.

a−b=3a - b = 3
a+b=−1a + b = -1
2a=22a = 2
a=1a = 1

Substitute the value of a into the equation a−b=3a - b = 3 to find the value of b.

1−b=31 - b = 3
b=−2b = -2

The resulting quadratic function is:

f(x)=x2−2x−3f(x) = x^2 - 2x - 3
Graph of f(x)=x2−2x−3f(x) = x^2 - 2x - 3
Parabola passing through points K(−1,0)K(-1, 0), L(0,−3)L(0, -3), and M(1,−4)M(1, -4).

Vertex Point

If we know the coordinates of the vertex (h, k) and another point (p, q) on the graph, we can use the vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k to find the value of a.

Example:

Find the quadratic function that has a vertex at (2, 0) and passes through the point (4, 4).

Use the vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k with h = 2 and k = 0.

f(x)=a(x−2)2+0=a(x−2)2f(x) = a(x - 2)^2 + 0 = a(x - 2)^2

Substitute the point (4, 4) to find the value of a.

4=a(4−2)24 = a(4 - 2)^2
4=4a4 = 4a
a=1a = 1

The resulting quadratic function is:

f(x)=(x−2)2=x2−4x+4f(x) = (x - 2)^2 = x^2 - 4x + 4
Graph of f(x)=(x−2)2f(x) = (x - 2)^2
Parabola with vertex at (2,0)(2, 0) and passing through point (4,4)(4, 4).

Equation Roots

If we know the roots (x-intercepts) p and q of the quadratic function, and an additional point (r, s) on the curve, we can use the factored form f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q).

Example:

Find the quadratic function that has roots at x = -2 and x = 3, and passes through the point (1, -6).

Use the factored form f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q) with p = -2 and q = 3.

f(x)=a(x−(−2))(x−3)=a(x+2)(x−3)f(x) = a(x - (-2))(x - 3) = a(x + 2)(x - 3)

Substitute the point (1, -6) to find the value of a.

−6=a(1+2)(1−3)-6 = a(1 + 2)(1 - 3)
−6=a(3)(−2)-6 = a(3)(-2)
−6=−6a-6 = -6a
a=1a = 1

The resulting quadratic function is:

f(x)=(x+2)(x−3)=x2−x−6f(x) = (x + 2)(x - 3) = x^2 - x - 6
Graph of f(x)=(x+2)(x−3)f(x) = (x + 2)(x - 3)
Parabola with roots at x=−2x = -2 and x=3x = 3, passing through point (1,−6)(1, -6).

Axis of Symmetry and Discriminant

We can also construct a quadratic function by knowing the axis of symmetry (or the x-coordinate of the vertex) and the value of the discriminant.

Example:

Find a quadratic function with axis of symmetry x = 1 and discriminant D = 16.

From the axis of symmetry x = 1, we know that −b2a=1-\frac{b}{2a} = 1, so b=−2ab = -2a.

From the discriminant D = 16, we know that b2−4ac=16b^2 - 4ac = 16.

Substitute b=−2ab = -2a into the discriminant equation.

(−2a)2−4ac=16(-2a)^2 - 4ac = 16
4a2−4ac=164a^2 - 4ac = 16
4a(a−c)=164a(a - c) = 16
a(a−c)=4a(a - c) = 4

There are many values of a and c that satisfy this equation. Let's take the simple case with a = 1.

1(1−c)=41(1 - c) = 4
1−c=41 - c = 4
c=−3c = -3

With a = 1, b = -2, and c = -3, the resulting quadratic function is:

f(x)=x2−2x−3f(x) = x^2 - 2x - 3

Symmetric Coordinates

We can construct a quadratic function by utilizing the symmetry property of parabolas.

Example:

Determine the quadratic function that passes through the points (0, 0), (4, 1), and (-4, 1).

Since the points (4, 1) and (-4, 1) have the same y-value and are at the same distance from the y-axis, the curve is symmetric about the y-axis. This means the axis of symmetry is x = 0, and the vertex is at (0, 0).

Use the vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k with h = 0 and k = 0.

f(x)=ax2f(x) = ax^2

Substitute the point (4, 1) to find the value of a.

1=a(4)21 = a(4)^2
1=16a1 = 16a
a=116a = \frac{1}{16}

The resulting quadratic function is:

f(x)=116x2f(x) = \frac{1}{16}x^2
Graph of f(x)=116x2f(x) = \frac{1}{16}x^2
Parabola passing through points (0,0)(0, 0), (4,1)(4, 1), and (−4,1)(-4, 1).

Transformations Between Quadratic Function Forms

Standard Form to Vertex Form

To convert f(x)=ax2+bx+cf(x) = ax^2 + bx + c to vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k:

  1. Determine the x-coordinate of the vertex: h=−b2ah = -\frac{b}{2a}
  2. Calculate the function value at the vertex: k=f(h)=f(−b2a)k = f(h) = f(-\frac{b}{2a})
  3. Or use the formula: k=c−b24ak = c - \frac{b^2}{4a}

Standard Form to Factored Form

To convert f(x)=ax2+bx+cf(x) = ax^2 + bx + c to factored form f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q):

  1. Determine the roots of the equation ax2+bx+c=0ax^2 + bx + c = 0 using the formula:

    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  2. If the roots are p and q, then:

    f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q)

Factored Form to Standard Form

To convert f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q) to standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c:

f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q)
f(x)=a(x2−qx−px+pq)f(x) = a(x^2 - qx - px + pq)
f(x)=a(x2−(p+q)x+pq)f(x) = a(x^2 - (p+q)x + pq)
f(x)=ax2−a(p+q)x+apqf(x) = ax^2 - a(p+q)x + apq

By comparing with the standard form, we get:

  • b=−a(p+q)b = -a(p+q)
  • c=apqc = apq

Exercises and Solutions

Exercise 1

Find the quadratic function that passes through the points (-2, 4), (1, -5), and (3, 7).

Answer:

Substitute the points into the standard equation:

4=a(−2)2+b(−2)+c=4a−2b+c4 = a(-2)^2 + b(-2) + c = 4a - 2b + c
−5=a(1)2+b(1)+c=a+b+c-5 = a(1)^2 + b(1) + c = a + b + c
7=a(3)2+b(3)+c=9a+3b+c7 = a(3)^2 + b(3) + c = 9a + 3b + c

From the second equation, c=−5−a−bc = -5 - a - b. Substitute into the first equation, then solve:

4=4a−2b+(−5−a−b)4 = 4a - 2b + (-5 - a - b)
9=3a−3b9 = 3a - 3b
3=a−b3 = a - b
a=3+ba = 3 + b

Substitute into the third equation:

7=9(3+b)+3b+(−5−(3+b)−b)7 = 9(3 + b) + 3b + (-5 - (3 + b) - b)
7=19+10b7 = 19 + 10b
b=−1210=−65b = -\frac{12}{10} = -\frac{6}{5}

Then:

a=3+(−65)=95a = 3 + (-\frac{6}{5}) = \frac{9}{5}
c=−5−95−(−65)=−285c = -5 - \frac{9}{5} - (-\frac{6}{5}) = -\frac{28}{5}

Therefore, the quadratic function is:

f(x)=95x2−65x−285f(x) = \frac{9}{5}x^2 - \frac{6}{5}x - \frac{28}{5}

Exercise 2

Determine the quadratic function that has a vertex at (-1, 4) and passes through the point (2, -5).

Answer:

Use the vertex form with h = -1 and k = 4:

f(x)=a(x−(−1))2+4=a(x+1)2+4f(x) = a(x - (-1))^2 + 4 = a(x + 1)^2 + 4

Substitute the point (2, -5):

−5=a(2+1)2+4-5 = a(2 + 1)^2 + 4
−5=9a+4-5 = 9a + 4
−9=9a-9 = 9a
a=−1a = -1

Therefore, the quadratic function is:

f(x)=−(x+1)2+4=−x2−2x+3f(x) = -(x + 1)^2 + 4 = -x^2 - 2x + 3

Exercise 3

Determine the quadratic function that has roots at x = -3 and x = 2, and has a maximum value of 4.

Answer:

Use the factored form with p = -3 and q = 2:

f(x)=a(x−(−3))(x−2)=a(x+3)(x−2)f(x) = a(x - (-3))(x - 2) = a(x + 3)(x - 2)

Since the function has a maximum value, a<0a < 0.

The vertex is at x=p+q2=−3+22=−12x = \frac{p + q}{2} = \frac{-3 + 2}{2} = -\frac{1}{2}.

Substitute x = -1/2 into the factored form and use the fact that the maximum value is 4:

f(−12)=a(−12+3)(−12−2)=−a254=4f(-\frac{1}{2}) = a(-\frac{1}{2} + 3)(-\frac{1}{2} - 2) = -a\frac{25}{4} = 4
a=−1625a = -\frac{16}{25}

Therefore, the quadratic function is:

f(x)=−1625(x+3)(x−2)=−1625x2−1625x+9625f(x) = -\frac{16}{25}(x + 3)(x - 2) = -\frac{16}{25}x^2 - \frac{16}{25}x + \frac{96}{25}
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