## anonymous 4 years ago Find the axis of symmetry for the following function. f(x) = -2x2 - 4x x=?

1. anonymous

hold on im gonna look in my notes

2. anonymous

well that was wrong. vertex is $-\frac{b}{2a}=-\frac{4}{4}=-1$

3. anonymous

how would i find the y-intercept of this problem?

4. anonymous

y = -2(x^2 + 2x) y = -2(x^2 + 2x +1 - 1) *divide the middle number by two then square it (subtract and add that number) y = -2 (x^2 + 2x + 1) +2 *take the subtracted number and multiply it by a to move it out of the brackets y = -2(x + 1) ^2 + 2 *take the middle term and divide it by 2 to get the x value. then square the whole bracket your vertex is (-1, 2) * h value is always opposite of the sign

5. anonymous

x = -1 <----- axis of symmetry as for your y - intercept set all your x values to 0 and solve for y y = -2x^2 - 4x y = -2(0)^2 - 4(0) y = 0 * your y intercept is 0 Note f(x) is y

6. anonymous

y intercept, set it equal to zero and solve get $-2x^2-4x=0$ $-2x(x+2)=0$ $x=0,x=-2$