# Quadratic Function + linear function

• Dec 20th 2011, 05:18 PM
gilagila
Write the following into the form of
$f(x)=a(x-h)^2+k$

and find the vertex, the symmetric axis and y-intercept

a) $f(x)=x^2+2x-1$
$f(x)=x^2+2x-1$
$x^2+2x+(2/2)^2=1+(2/2)^2$
$(x+1)^2=2$
$f(x)=(x+1)^2-2$

Vertex = (-1,-2), Symmetric, x=-1, y-intercept, when x=0, y=-1

Am i doing the correct method to calculate ??

b) $f(x)=-x^2+4x-1$
$f(x)=-[x^2-4x+1]$
$-[x^2-4x+(-4/2)^2]=-1+(-4/2)$
$-[(x-2)^2]=-3$
$f(x)=-(x-2)^2+3$

Vertex = (2,3), symmetric x=2, y-intercept=-1

c) $f(x)=3x^2+6x-3$
$f(x)=3[x^2+2x-1]$
$3[x^2+2x+(2/2)^2]=1+(2/2)^2$
$3[(x+1)^2]=2$
$f(x)=3(x+1)^2-6$

Vertex (-1,-6), symmetrix =-1, y-intercept=-6

am i correct ?
• Dec 20th 2011, 05:29 PM
Prove It
Re: Quadratic Function + linear function
Quote:

Originally Posted by gilagila
Write the following into the form of
$f(x)=a(x-h)^2+k$

and find the vertex, the symmetric axis and y-intercept

a) $f(x)=x^2+2x-1$
$y=x^2+2x-1$
$1=x^2+2x$
$1+1^2=x^2+2x+1^2$
$2=(x+1)^2$
$y=(x+1)^2-2$
$f(x)=(x+1)^2-2$

Vertex = (1,-2), Symmetric, x=1, y-intercept, when x=0, y=-1

Am i doing the correct method to calculate ??

b) $f(x)=-x^2+4x-1$
$f(x)=x^2-4x+1$
$-1=x^2-4x$
$-1+1^2=x^2-4x+1^2$
$(x+1)^2=0$
then what is the next step ? the answer is $f(x)=-(x-2)^2+3$

same goes to

c) $f(x)=3x^2+6x-3$
$f(x)=x^2+2x-1$
$1=x^2+2x$
$1+1^2=x^2+2x+1^2$
$2=(x+1)^2$
[tex]f(x)=(x+1)^2-2

answer is $f(x)=3(x+1)^2-6$

You can't factorise and then get rid of the other factors, and then say the functions are equal.

For the second, you should be writing

\displaystyle \begin{align*} f(x) &= -x^2 + 4x - 1 \\ &= -\left(x^2 - 4x + 1\right) \\ &= -\left[x^2 - 4x + (-2)^2 - (-2)^2 + 1\right] \\ &= -\left[\left(x - 2\right)^2 - 4 + 1\right] \\ &= -\left[(x - 2)^2 - 3\right] \\ &= -(x - 2)^2 + 3 \end{align*}

See what you can do for the third.
• Dec 20th 2011, 05:36 PM
gilagila
Re: Quadratic Function + linear function
Quote:

Originally Posted by Prove It
You can't factorise and then get rid of the other factors, and then say the functions are equal.

For the second, you should be writing

\displaystyle \begin{align*} f(x) &= -x^2 + 4x - 1 \\ &= -\left(x^2 - 4x + 1\right) \\ &= -\left[x^2 - 4x + (-2)^2 - (-2)^2 + 1\right] \\ &= -\left[\left(x - 2\right)^2 - 4 + 1\right] \\ &= -\left[(x - 2)^2 - 3\right] \\ &= -(x - 2)^2 + 3 \end{align*}

See what you can do for the third.

oops, just notice that my completing in square is wrong just now, able to correct, get the same answer as you.

how about the no.3, i make some changes, is it correct ?
• Dec 20th 2011, 05:41 PM
Quacky
Re: Quadratic Function + linear function
Quote:

Originally Posted by gilagila
Write the following into the form of
$f(x)=a(x-h)^2+k$

and find the vertex, the symmetric axis and y-intercept

a) $f(x)=x^2+2x-1$
$y=x^2+2x-1$
$1=x^2+2x$
$1+1^2=x^2+2x+1^2$
$2=(x+1)^2$
$y=(x+1)^2-2$
$f(x)=(x+1)^2-2$

Vertex = (-1,-2), Symmetric, x=-1, y-intercept, when x=0, y=-1. Q:Note this correction.

b) $f(x)=-x^2+4x-1$
Q:Your working is incorrect. $-x^2+4x+1\neq x^2-4x-1$
Starting again:
$-x^2+4x-1$
$=-[x^2-4x+1]$
$=-[(x-2)^2-2^2+1]$
$=-(x-2)^2+3$
Now solve for the vertices, symmetry etc.

same goes to

c) $f(x)=3x^2+6x-3$

Q:Again, incorrect. $\frac{f(x)}{3}\neq{f(x)}}$ - the only time you can divide through by a common factor is when identifying roots.

$f(x)=3[x^2+2x-1]$

$=3[(x+1)^2-1^2-1]$

$=3[(x+1)^2-2]$

$=3(x+1)^2-6$

Note the corrections.
• Dec 20th 2011, 05:46 PM
gilagila
Re: Quadratic Function + linear function
Quote:

Originally Posted by Quacky
Note the corrections.

oops, never notice for Q3 the k also need x3, ya..finally resolved all the question.

TQ