Determining the Equation of a Graph0%

Polynomials & Quadratics · Topic 6 of 10

Determining the Equation of a Graph

Video lesson · from 34:102 worked examples

One lesson video covers all of Polynomials & Quadratics, so it opens at 34:10 for this topic — not from the beginning.

Theory

Given the roots and at least one other point on the graph, we can establish the graph's equation.

If the roots are x=a,x=bx = a, x = b and x=cx = c, the equation is of the form y=k(xa)(xb)(xc)y = k(x-a)(x-b)(x-c). We can find kk using the other given point.

Repeated Roots

If a repeated root exists, then a stationary point lies on the x-axis.

Recall that a repeated root exists when two roots, and hence two factors, are equal.

If a graph has a root x=ax=a and a repeated root at x=bx=b (where it touches the x-axis), the equation is of the form y=k(xa)(xb)2y = k(x-a)(x-b)^2.

⚠️ Common Examiner Traps

  • You must find kk. Many candidates find the factors and then stop, with no strategy at all for finding k. Substitute another point from the graph — usually the yy-intercept — and solve.
  • Signs of the roots: sign errors here are extremely common. A root at x=3x = -3 gives the factor (x+3)(x+3).
  • A touch means a repeated factor: where the curve touches the axis and turns back, that factor is squared. Treating it as a single root gives a curve of the wrong degree.
  • Count the degree: a cubic needs three factors in total, counting the repeat twice.

Worked examples

Example 1

Find the equation of the cubic shown in the diagram:

-6-31(0, -36)

The roots are x=6,x=3x = -6, x = -3 and x=1x = 1.

So the equation is of the form:

y=k(x+6)(x+3)(x1)y = k(x+6)(x+3)(x-1)

We are given the y-intercept (0,36)(0, -36). Substitute x=0x=0 and y=36y=-36:

36=k(0+6)(0+3)(01)-36 = k(0+6)(0+3)(0-1)
36=k(6)(3)(1)-36 = k(6)(3)(-1)
36=18k-36 = -18k
k=2k = 2

Therefore, the valid equation is:

y=2(x+6)(x+3)(x1)y = 2(x+6)(x+3)(x-1)

Example 2

Repeated Roots. Find the equation of the cubic shown in the diagram:

(-2, 0)(0, 9)(3, 0)

There is a root at x=2x = -2 and a repeated root (touches the x-axis) at x=3x = 3.

The equation is of the form:

y=k(x+2)(x3)2y = k(x+2)(x-3)^2

We use the point (0,9)(0, 9) to find kk. Substitute x=0,y=9x=0, y=9:

9=k(0+2)(03)29 = k(0+2)(0-3)^2
9=k(2)(3)29 = k(2)(-3)^2
9=k(2)(9)9 = k(2)(9)
9=18k9 = 18k
k=12k = \frac{1}{2}

Therefore, the Valid equation is:

y=12(x+2)(x3)2y = \frac{1}{2}(x+2)(x-3)^2