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find the points on the curve y = x^3 + 3x^2 - 9x + 3 where the tangent …

Question

find the points on the curve y = x^3 + 3x^2 - 9x + 3 where the tangent is horizontal. smaller x - value (x,y)=() larger x - value (x,y)=() 20. -/2 points draw a diagram to show that there are two tangent lines to the parabola of y = x^2 that pass through the point (0, - 25). find the coordinates of the points where these tangent lines intersect the parabola. smaller x - value (x,y)=() larger x - value (x,y)=()

Explanation:

Step1: Differentiate the function

The derivative of $y = x^{3}+3x^{2}-9x + 3$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $y'=3x^{2}+6x - 9$.

Step2: Set the derivative equal to zero

Since the tangent is horizontal when the slope (derivative) is zero, we set $3x^{2}+6x - 9 = 0$. Divide through by 3 to get $x^{2}+2x - 3=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}+2x - 3=(x + 3)(x - 1)=0$. Then, $x=-3$ or $x = 1$.

Step4: Find the corresponding y - values

When $x=-3$, $y=(-3)^{3}+3(-3)^{2}-9(-3)+3=-27 + 27+27 + 3=30$.
When $x = 1$, $y=1^{3}+3\times1^{2}-9\times1 + 3=1 + 3-9 + 3=-2$.

Answer:

smaller $x$-value $(x,y)=(-3,30)$
larger $x$-value $(x,y)=(1,-2)$