algebra help
like how in y = mx + b b = y-intercept in y= -4x^2 + 8x + 3 ^ would be the y-intercept
vertex form... ... ... y = m(x-h)^2 + k y = -x^2 + 2x + 5 = -(x^2 - 2x - 5) = -(x^2 - 2x + 1 - 1 - 5) = -(x^2 - 2x + 1) + 6 = -(x-1)^2 + 6 ^ ^ so (1, 6)
crosses the x-axis at x = -6 and x= 2 because slope is 1/2 it is fat
slope is 1/2 therefore it is fatter
vertex is (3, 6)
nope vertex is (3, 6)
A would be correct :)
A
because vertex is (3, -6) and plugged in point (0, 0)
f(x) = x^2 - 6x - 5 = x^2 - 6x + 9 - 9 - 5 = x^2 - 6x + 9 - 14 = (x - 3)^2 - 14
-1/3 , 5
y = 12x^2 + 16x - 3 factor... = (2x + 3) (6x - 1) solve for each when 2x+3=0 and 6x-1=0 x = -2/3 and x = 1/6
no real solutions... so idk :/
also no real solutions... you can get an account on wolframalpha and they'll show you the steps if you want... but idk :/ sorry http://www.wolframalpha.com/input/?i=0+%3D+5x%5E2+-+6x+%2B+2
does not factor out evenly... so D
sure :)
k :)
I gtg :/ I'll be back later later maybe...
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