How do you find the vertex and intercepts for #y = 10x – 3x^2#?

(y)-intercept: Set (x = 0) in the equation (y = 10x - 3x^2) and solve for (y). This gives you the value of (y) when (x = 0), which is the (y)-intercept.
(x)-intercepts: Set (y = 0) in the equation (y = 10x - 3x^2) and solve for (x). This gives you the values of (x) where the graph intersects the (x)-axis, which are the (x)-intercepts.

Question

Ethan Adkins · Accepted Answer

The vertex is #(5/3,25/3)#
The intercepts are (0,0) and #(10/3,0)# you can find the coordinate x of the vertex using the formula #x=-b/(2a)# where a is the coefficient of #x^2# and b of x Since a=-3 and b=10, you have #x=-10/-6=5/3# Then you can put this value in x and find y #y=10(5/3)-3(5/3)^2# #y=50/3-3(25/9)# #y=50/3-25/3# #y=25/3# One of the intercepts is the origin (0,0) because y has no known term By solving the equation #10x-3x^2=0# you can find the other one: #x(10-3x)=0# #10-3x=0# #x=10/3#

Eva Abramson · Answer

undefined To find the vertex and intercepts for the equation $y = 10x - 3x^2$:

1. **Vertex**: The vertex of a quadratic function in the form $y = ax^2 + bx + c$ can be found using the formula $x = \frac{-b}{2a}$. In this case, $a = -3$ and $b = 10$. Plug these values into the formula to find the $x$-coordinate of the vertex. Once you have the $x$-coordinate, substitute it back into the equation to find the corresponding $y$-coordinate.

2. **Intercepts**:
   - **$y$-intercept**: Set $x = 0$ in the equation $y = 10x - 3x^2$ and solve for $y$. This gives you the value of $y$ when $x = 0$, which is the $y$-intercept.
   - **$x$-intercepts**: Set $y = 0$ in the equation $y = 10x - 3x^2$ and solve for $x$. This gives you the values of $x$ where the graph intersects the $x$-axis, which are the $x$-intercepts.