How do you write the quadratic in vertex form given vertex is (1/10, -9/10) and y intercept is -1?

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Answer 1 · 2015-05-15T23:49:55

Vertex form:#f(x) = a(x -a/2b)^2 + f(-a/2b)#
Standard form f(x) = ax^2 + bx + c
First find a, b, and c. We have as information:

x of vertex:# (-b/2a) = 1/10 -> a = (-10b)/2 = -5b# (1)

y intercept (-1) gives -> c = -1

y of vertex:# f(-b/2a) = (-9/10) = a/100 + b/10 - 1 # ->

a + 10b - 100 + 90 = 0 -> a + 10b - 10 = 0 . Replace a by -5b

-5b + 10b = 10 -> b = 10/5 = 2 -> a = -10.

Standard form: #f(x) = -10x^2 + 2x - 1#

Finally, vertex form: #f(x) = -10(x - 1/10)^2 - 90/100#

Check: Develop f(x)
#f(x) = -10(x^2 - (2x)/10 + 1/100) - 90/100 =#
#= -10x^2 + 2x - 10/100 - 90/100 = - 10x^2 + 2x - 1 #. Correct