Math Snap

STEP 1

What is this asking?
We're going to explore this cool function $y = -x^4 + 4x^3$, find its intercepts, figure out where it's increasing and decreasing, where it's curving up or down, and then draw a sweet sketch of it!
Watch out!
Don't mix up increasing/decreasing with concave up/concave down.
They're related, but totally different things!
Also, remember that relative extrema are like peaks and valleys, and inflection points are where the curve changes its bendiness.

STEP 2

1. Find the intercepts
2. Analyze increasing/decreasing behavior and extrema
3. Analyze concavity and inflection points
4. Sketch the graph

STEP 3

To find the y-intercept, we set $x = 0$ in our function: $y = -(0)^4 + 4(0)^3 = 0$.
So, the y-intercept is at the point $(0, 0)$.
Boom!

STEP 4

For the x-intercepts, we set $y = 0$ and solve for $x$:
$$0 = -x^4 + 4x^3$$ $$0 = x^3(-x + 4)$$This equation is true if $x^3 = 0$ or if $-x + 4 = 0$.
So, our x-intercepts are $x = 0$ and $x = 4$, giving us the points $(0, 0)$ and $(4, 0)$.
Awesome!

STEP 5

First, we need the first derivative of our function, which tells us about the slope:
$$y' = -4x^3 + 12x^2$$

STEP 6

To find where the function is increasing or decreasing, we set the first derivative equal to zero and solve for $x$:
$$0 = -4x^3 + 12x^2$$ $$0 = -4x^2(x - 3)$$So, $x = 0$ and $x = 3$ are our critical points.

STEP 7

Now, we test values around these critical points to see if the derivative is positive (increasing) or negative (decreasing).
Let's pick $x = -1$, $x = 1$, and $x = 4$.
• For $x = -1$, $y' = -4(-1)^3 + 12(-1)^2 = 4 + 12 = 16 > 0$, so the function is increasing before $x = 0$.
• For $x = 1$, $y' = -4(1)^3 + 12(1)^2 = -4 + 12 = 8 > 0$, so the function is increasing between $x = 0$ and $x = 3$.
• For $x = 4$, $y' = -4(4)^3 + 12(4)^2 = -256 + 192 = -64 < 0$, so the function is decreasing after $x = 3$.

STEP 8

Since the function goes from increasing to decreasing at $x = 3$, we have a relative maximum there.
Let's find the y-value: $y = -(3)^4 + 4(3)^3 = -81 + 108 = 27$.
So, the relative maximum is at $(3, 27)$.

STEP 9

We need the second derivative to talk about concavity:
$$y'' = -12x^2 + 24x$$

STEP 10

We set the second derivative equal to zero and solve for $x$:
$$0 = -12x^2 + 24x$$ $$0 = -12x(x - 2)$$So, $x = 0$ and $x = 2$ are our potential inflection points.

STEP 11

Let's test $x = -1$, $x = 1$, and $x = 3$:
• For $x = -1$, $y'' = -12(-1)^2 + 24(-1) = -12 - 24 = -36 < 0$, so concave down.
• For $x = 1$, $y'' = -12(1)^2 + 24(1) = -12 + 24 = 12 > 0$, so concave up.
• For $x = 3$, $y'' = -12(3)^2 + 24(3) = -108 + 72 = -36 < 0$, so concave down.

STEP 12

Since the concavity changes at both $x = 0$ and $x = 2$, these are inflection points.
We already know the y-value for $x = 0$ is $0$.
For $x = 2$, $y = -(2)^4 + 4(2)^3 = -16 + 32 = 16$.
So, our inflection points are $(0, 0)$ and $(2, 16)$.

STEP 13

Using all this info, we can sketch the graph!
It starts increasing, passes through $(0,0)$, keeps increasing and changes from concave down to concave up at $(0,0)$, reaches an inflection point at $(2,16)$, hits a maximum at $(3,27)$, then decreases, passing through $(4,0)$, and continues decreasing while being concave down.

The x-intercepts are $(0, 0)$ and $(4, 0)$, and the y-intercept is $(0, 0)$.
The function is increasing on $(-\infty, 3)$ and decreasing on $(3, \infty)$, with a relative maximum at $(3, 27)$.
It's concave up on $(0, 2)$ and concave down on $(-\infty, 0)$ and $(2, \infty)$, with inflection points at $(0, 0)$ and $(2, 16)$.