Step-by-step explanation:
y=(x+5)(x-1)
y=-5 x=0
y=0. x=-5
y=0 x=1
now you got the points so graph will be like
Zeroes:
We must solve

To do so, we define the auxiliary variable
. The equation becomes

The quadratic formula yields the solutions

Substituting back
gives

So, the zeroes are -6, -3, 3, 6.
Turning points:
Turning points are points where a function stops being increasing to become decreasing, or vice versa. Since functions are increasing when their first derivative is positive and decreasing when it's negative, turning points are points where the first derivative is zero.
We have

If we set the derivative to be zero, we have

So, the derivative is zero if x=0 or

Answer:
x intercept. The x intercept is the point where the line crosses the x axis. At this point y = 0. y intercept. The y intercept is the point where the line crosses the y axis.
Answer:
option c is correct.
Step-by-step explanation:
![7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{16x}\right)-3\left(\sqrt[3]{8x}\right)](https://tex.z-dn.net/?f=7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B16x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B8x%7D%5Cright%29)
WE need to simplify this equation.
Solve the parenthesis of each term.
![=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right](https://tex.z-dn.net/?f=%3D7%5Cleft%5Csqrt%5B3%5D%7B2x%7D%5Cright-3%5Cleft%5Csqrt%5B3%5D%7B16x%7D%5Cright-3%5Cleft%5Csqrt%5B3%5D%7B8x%7D%5Cright)
Now, We will find factors of the terms inside the square root
factors of 2: 2
factors of 16 : 2x2x2x2
factors of 8: 2x2x2
Putting these values in our equation:![=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2X2 x}\right)-3\left(\sqrt[3]{2X2X2 x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2X2X2} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3] {2 x}\right)-3\left(\sqrt[3]{2^3} \sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2x}\right)-3*2\left(\sqrt[3] {2 x}\right)-3*2\left(\sqrt[3]{x}\right)\\=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)](https://tex.z-dn.net/?f=%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2X2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%20x%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%7D%20%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2X2X2%7D%20%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2%5E3%7D%20%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%5Cleft%28%5Csqrt%5B3%5D%7B2%5E3%7D%20%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2x%7D%5Cright%29-3%2A2%5Cleft%28%5Csqrt%5B3%5D%20%7B2%20x%7D%5Cright%29-3%2A2%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%5Cright%29-6%5Cleft%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29)
Adding like terms we get:
![=7\left(\sqrt[3]{2}\sqrt[3]{x}\right)-6\left(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right\\=(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\](https://tex.z-dn.net/?f=%3D7%5Cleft%28%5Csqrt%5B3%5D%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%5Cright%29-6%5Cleft%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%5C%5C%3D%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5C)
![(\sqrt[3] {2}\sqrt[3]{x})-6\left(\sqrt[3]{x}\right)\\can\,\,be \,\, written\,\, as\,\,\\(\sqrt[3] {2x})-6\left(\sqrt[3]{x}\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%20%7B2%7D%5Csqrt%5B3%5D%7Bx%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%5C%5Ccan%5C%2C%5C%2Cbe%20%5C%2C%5C%2C%20written%5C%2C%5C%2C%20as%5C%2C%5C%2C%5C%5C%28%5Csqrt%5B3%5D%20%7B2x%7D%29-6%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29)
So, option c is correct
What’s your question??
Hope I helepd