Porisma

Worked solution

Recover a cubic from its graph

A cubic polynomial curve, CC, is shown below, find a possible equation for CC.

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Source cubic grid: x-axis labelled from −4 to 6; y-axis levels −2,−1,0,1. Curve crosses x at −3,2,5 and y at −2, rises to the left and falls to the right.
Source cubic grid: x-axis labelled from −4 to 6; y-axis levels −2,−1,0,1. Curve crosses x at −3,2,5 and y at −2, rises to the left and falls to the right.
Source cubic grid: x-axis labelled from −4 to 6; y-axis levels −2,−1,0,1. Curve crosses x at −3,2,5 and y at −2, rises to the left and falls to the right.

CJC · Lecture · Example 7

Answer

y=−115(x+3)(x−2)(x−5)y=-\frac1{15}(x+3)(x-2)(x-5).

Use the three crossings to choose the factors

The source graph crosses the xx-axis at −3-3, 22 and 55. Since it is a cubic, use one factor for each crossing:

f(x)=a[x−(−3)](x−2)(x−5)=a(x+3)(x−2)(x−5).\begin{aligned}&f(x)\\ & =a\big[x-(-3)\big](x-2)(x-5)\\ & =a(x+3)(x-2)(x-5).\end{aligned}

The constant aa sets the vertical scale and direction; the roots alone do not determine it.

How did this part go?

Use the y-intercept to determine the scale

Answer

a=−1/15a=-1/15.

Both source copies give f(0)=−2f(0)=-2 and the complete substitution:

a(0+3)(0−2)(0−5)=−2,30a=−2,a=−115.a(0+3)(0-2)(0-5)=-2,\qquad30a=-2,\qquad a=-\frac1{15}.

Hence f(x)=−115(x+3)(x−2)(x−5)f(x)=-\frac1{15}(x+3)(x-2)(x-5). Independently this gives all three printed zeros and f(0)=−2f(0)=-2; the negative leading coefficient also matches the graph’s upward left end and downward right end.

How did this part go?

Common mistakes

Do not set the leading coefficient to −1 merely because the curve falls on the right. Use the y-intercept to find its magnitude.

Solution notes

Prompt graph is on PDF/printed page 16; the complete solution continues on page 17 in both copies. Grid labels x=−4,…,6 and y=−2,…,1 are retained in the mathematical reconstruction. No printed marks or external attribution.

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