Power functions and first sketches, question 16: UBC MATH 100 practice

Question 16 of 17Capstone5 marks · 9 minTheory UBC Math 100 §1.2Key results

Let $y = kx^{2} + x^{3}$ , where $k$ is a nonzero constant. Which statement about the graph is correct for both signs of $k$ ?

The graph meets the x-axis at the origin and at x = −k. Near the origin it looks like y = x³, passing straight through it, and far from the origin it follows y = kx², rising on both sides if k > 0 and falling on both sides if k < 0.
The graph touches the x-axis at the origin without crossing it and crosses it once more, at x = k. Near the origin it looks like y = kx², and far from the origin it looks like y = x³, falling on the left and rising on the right.
The graph crosses the x-axis at the origin, changing sign there, and crosses it again at x = −k. Near the origin it looks like y = kx², and far from the origin it looks like y = x³, falling on the left and rising on the right.
The graph touches the x-axis at the origin without crossing it and crosses it once more, at x = −k. Near the origin it looks like the parabola y = kx², opening up if k > 0 and down if k < 0, and far from the origin it looks like y = x³, falling on the left and rising on the right.
The graph touches the x-axis at the origin and crosses it at x = −k. Near the origin it looks like y = kx², and far from the origin it keeps following y = kx², so when k < 0 it falls on both sides.

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Theory UBC Math 100 §1.2Key results

Worked solution

Strategy

Treat k as a fixed number and run the usual sketch: zeros from factoring, the dominant term near zero, the dominant term far away. Then ask which conclusions change when k changes sign, and what the repeated factor x² does at the origin.

Working

$kx^{2} + x^{3} = x^{2}(x + k) = 0 \iff x = 0 \text{ or } x = -k$
Factor out the lowest power. For every nonzero k there are exactly two zeros, and the second sits on the opposite side of 0 from k.
$x^{2} \geq 0 \quad\Longrightarrow\quad y \text{ has the sign of } x + k \text{ for } x \neq 0$
So y changes sign at x = −k, where the graph crosses, but not at the origin, where it only touches the axis.
$\text{Near } 0:\quad y \approx kx^{2}$
The lower power dominates near the origin: an upward bowl if k > 0, a downward one if k < 0.
$\text{Large } |x|:\quad y \approx x^{3}$
The higher power dominates far away and carries no k, so the graph falls on the far left and rises on the far right for either sign.

Answer

y = x^{2}(x + k):\ \text{touches at } 0,\ \text{crosses at } x = -k;\quad y \approx kx^{2} \text{ near } 0,\quad y \approx x^{3} \text{ far away}

Check

With k = 2, y(−3) = −9 and y(−1) = 1, a crossing at −2, while y(±0.1) > 0; with k = −2, y(1) = −1 and y(3) = 9.

Where this goes wrong

Treating the origin as an ordinary crossing. The zero at 0 comes from the factor x², which never changes sign, so the graph only touches the axis there; the one crossing is at x = −k.

The idea behind it

The power functions and first sketches lesson, a figure to move, in a few steps

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Power functions and first sketches practice question 16