## anonymous one year ago Find the maximum or minimum of the following quadratic function: y = -2x2 + 4x - 1. -1 0 1 4

1. anonymous

2. campbell_st

is this question algebra of calculus...?

3. campbell_st

ok... here is the the key, the parabola is concave down, since the coefficient of x^2 is -2 so for a quadratic $ax^2 + bx + c = 0$ the line of symmetry is $x = \frac{-b}{2 \times a}$ you have a = -2 and b = 4 the maximum of minimum lies on the line of symmetry so find the value of x, and then substitute it into the original equation to find the max value

4. anonymous

I don't quite understand... I just need to know whether its a b c or d im learning how to do it I just don't have the time...

5. campbell_st

yes but to find it, a simple method is to find the line of symmetry and the substitute that value into the equation

6. campbell_st

and alternate method is to just graph the curve and read off the value.

7. anonymous

look im not like most people on here I know nearly nothing about math which is why im taking this class.. I don't have any idea what youre talking about and im stressing about this cause I need to pass...

8. campbell_st

ok... then use this site https://www.desmos.com/calculator type in the equation on the left and then read off the highest y value

9. anonymous

1?...

10. campbell_st

that correct.... the line of symmetry is x = -4/(2 x -2) so x = 1 subsitute that value into the equation and you get y = -2 *(1)^2 + 4*(1) - 1 or y = 1 so the parabola is concave down, the you have a maximum at y = 1

11. anonymous

so one is the answer?

12. campbell_st

that's correct