Constructing a quadratic function means determining f(x)=ax2+bx+c from the given information. Known points lead to a system for the coefficients, known roots lead to factored form, and a known vertex leads to vertex form.
Three points with distinct x-coordinates determine exactly one polynomial of degree at most 2. It is a quadratic function only when the resulting coefficient satisfies a=0.
If we have three points (x1,y1), (x2,y2), and (x3,y3), we can substitute these values into the standard equation to get three linear equations with three variables a, b, and c.
y1=ax12+bx1+c
y2=ax22+bx2+c
y3=ax32+bx3+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)+c
L(0,−3):−3=a(0)2+b(0)+c
M(1,−4):−4=a(1)2+b(1)+c
Simplify these equations.
0=a−b+c
−3=c
−4=a+b+c
Substitute c=−3 into the first and third equations.
0=a−b−3
−4=a+b−3
Simplify and add the two equations to find the value of a.
a−b=3
a+b=−1
2a=2
a=1
Substitute the value of a into the equation a−b=3 to find the value of b.
1−b=3
b=−2
The resulting quadratic function is:
f(x)=x2−2x−3
Graph of f(x)=x2−2x−3
Parabola passing through points K(−1,0), L(0,−3), and M(1,−4).
If we know the roots (x-intercepts) p and q of the quadratic function, and an additional point (r,s) on the curve with r=p, r=q, and s=0, we can use the factored form f(x)=a(x−p)(x−q) to determine a.
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) with p=−2 and q=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(3)(−2)
−6=−6a
a=1
The resulting quadratic function is:
f(x)=(x+2)(x−3)=x2−x−6
Graph of f(x)=(x+2)(x−3)
Parabola with roots at x=−2 and x=3, passing through point (1,−6).
The axis of symmetry and the discriminant usually allow several quadratic functions. Another piece of information is needed to select one unique function.
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 −2ab=1, so b=−2a.
From the discriminant D=16, we know that b2−4ac=16.
Substitute b=−2a into the discriminant equation.
(−2a)2−4ac=16
4a2−4ac=16
4a(a−c)=16
a(a−c)=4
There are many values of a and c that satisfy this equation, so the data do not determine a unique function. To obtain one example, choose a=1.
1(1−c)=4
1−c=4
c=−3
With a=1, b=−2, and c=−3, the resulting quadratic function is:
Symmetric points can also determine a quadratic function.
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+k with h=0 and k=0.
f(x)=ax2
Substitute the point (4,1) to find the value of a.
1=a(4)2
1=16a
a=161
The resulting quadratic function is:
f(x)=161x2
Graph of f(x)=161x2
Parabola passing through points (0,0), (4,1), and (−4,1).