y=2x^3√9-x^2

find x and y intercept, i know we have x (a,0) y (0,b)

Ok, you can do the y intercept (x is zero).

for the x intercept, set y to zero.

0=x^2(2xsqrt9-1)
so you have a double zero at x=0, and then one more at
(2xsqrt9=1
or x= 1/(2sqrt9)

I did not get this, i did 0=2x^3*square root of 9-x^2, then I 2x^3(3-x) 6x^3-2x^4, 0=2x^3(x-3), but from here im lost for x intercept y intercept is easy.

To find the x-intercept, we set y = 0 and solve for x. Similarly, to find the y-intercept, we set x = 0 and solve for y.

Let's start with finding the x-intercept:

Setting y = 0 in the equation y = 2x^3√(9 - x^2), we have:
0 = 2x^3√(9 - x^2)

Since the square root (√) of a number cannot be negative, we can divide both sides of the equation by 2x^3:

0 / (2x^3) = √(9 - x^2)

Simplifying the equation gives:
0 = √(9 - x^2)

To solve for x, we square both sides of the equation:

0^2 = (√(9 - x^2))^2
0 = 9 - x^2

Rearranging the equation, we get:
x^2 = 9

Now, taking the square root of both sides, we have:

√(x^2) = √(9)
x = ±3

Therefore, the x-intercepts are at x = -3 and x = 3.

Now, let's find the y-intercept:

Setting x = 0 in the equation y = 2x^3√(9 - x^2), we have:
y = 2(0)^3√(9 - (0)^2)
y = 0√(9)
y = 0

Therefore, the y-intercept is at y = 0.

In summary:
- The x-intercepts are at x = -3 and x = 3.
- The y-intercept is at y = 0.