Question

complete the square to write the quadratic equation c(x)= x^2-3/4x+1 in vertex form

Answer 1

is it
\[\large c(x)=x^2-\frac{3}{4}x+1\]?

Answer 2

yes!

Answer 3

lets just find the vertex instead, how about that?

Answer 4

the first coordinate of the vertex is \(-\frac{b}{2a}\) which in your case is
\[-\frac{-\frac{3}{4}}{2}=\frac{3}{8}\]

Answer 5

so it is going to be
\[c(x)=(x-\frac{3}{8})^2+k\]
to find \(k\) replace \(x\) by \(\frac{3}{8}\) in the original expression, in other words compute
\[c(\frac{3}{8})\]

Answer 6

would the coordinates of the vertex be (.375, 1.421875)

Answer 7

lets skip decimals and use fractions instead

Answer 8

so (3/8, 91/64) ?

Answer 9

hmmm i think the 91 is wrong

Answer 10

\[\left(\frac{3}{8}\right)^2-\frac{3}{4}\times \frac{3}{8}+1\] is the first step

Answer 11

you should get
\[\frac{55}{64}\] i believe

Answer 12

my calculator is still spitting out 91/64 ok so (3/8)^2 = .140625
-3/4(3/8)= -.28125
.140625 + 1 -.28125= 1.421875 or 91/64

Answer 13

did I do anything incorrectly

Answer 14

\[\frac{9}{64}-\frac{9}{32}+1\\
\frac{9}{64}-\frac{18}{64}+1\\
\frac{-9}{64}+1\\
\frac{55}{64}\]

Answer 15

Oh okay, and then after we find the vertex what do we do?

Answer 16

then write it in vertex form
\[c(x)=\left(x-\frac{3}{8}\right)^2+\frac{55}{64}\]

Answer 17

Thanks!! I understand this now!!