Question

1. - / 4.54 points

find the points on the curve $y = 2x^3 + 3x^2 - 12x + 7$ where the tangent line is horizontal.
$(x, y) = \left(\quad\
ight)$ (smaller $x$-value)
$(x, y) = \left(\quad\
ight)$ (larger $x$-value)
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1. - / 4.54 points

find an equation of the tangent line to the curve $y = x\sqrt{x}$ that is parallel to the line $y = 9 + 3x$.
$y = \quad$
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Explanation:

Step1: Compute derivative of y

$y = 2x^3 + 3x^2 - 12x + 7$, so $y' = 6x^2 + 6x - 12$

Step2: Set derivative to 0 (horizontal tangent)

$6x^2 + 6x - 12 = 0 \implies x^2 + x - 2 = 0 \implies (x+2)(x-1)=0$
Solutions: $x=-2$ (smaller), $x=1$ (larger)

Step3: Find y for x=-2

$y=2(-2)^3 +3(-2)^2 -12(-2)+7=2(-8)+3(4)+24+7=-16+12+24+7=27$

Step4: Find y for x=1

$y=2(1)^3 +3(1)^2 -12(1)+7=2+3-12+7=0$

Step5: Rewrite y=x√x as power function

$y=x\sqrt{x}=x^{3/2}$, derivative $y'=\frac{3}{2}x^{1/2}$

Step6: Set derivative to 3 (parallel slope)

$\frac{3}{2}\sqrt{x}=3 \implies \sqrt{x}=2 \implies x=4$

Step7: Find y at x=4

$y=4\sqrt{4}=4*2=8$

Step8: Write tangent line equation

Using point-slope form: $y-8=3(x-4) \implies y=3x-12+8=3x-4$

Answer:

For Q19:
$(x,y)=(-2,27)$ (smaller x-value)
$(x,y)=(1,0)$ (larger x-value)
For Q20:
$y=3x-4$