Question

How do you find the vertex and the intercepts for `y = -5x^2 + 10x - 7`?

Answer 1

`V (1:-2)`

one intercept `(0;-7)`

Explanation:

You can find the vertex by applying the following formula:

`V_x=-b/(2a)`

and then substitute its value in the equation to find `V_y`:

#V_x=-10/(2(-5))=1

`V_y=-5*1^2+10*1-7=-2`

Then the vertex is `V (1:-2)`

To find the intercepts with y-axis, you would substitute `x=0` in the equation and, with x-axis, `y=0`:

`x=0 -> y=-7`

`y=0->-5x^2+10x-7=0->x=(-5+-sqrt(25-35))/-5->`there are no real solutions
graph{-5x^2+10x-7 [-2,3, -10, 1]}