# Function notation

• Dec 8th 2011, 04:09 PM
daword177
Function notation
Find the equation of the line. Write the equation using function notation.
Through (-2,4); perpendicular to f(x)=-2x+3

The answer is 1/2x +5 but I dont know how they got that.

2nd question

Through (6,-6); parallel to 5x+6y=18

the answer is -5/6x-1

I got y=-5/6x-1

My teacher gave me the cheat sheet for the review to make sure when I study that I can correct them.
• Dec 8th 2011, 04:10 PM
daword177
Re: Function notation
If you can please show your work so I can learn from it please! Thank you!
• Dec 8th 2011, 04:14 PM
Prove It
Re: Function notation
Quote:

Originally Posted by daword177
Find the equation of the line. Write the equation using function notation.
Through (-2,4); perpendicular to f(x)=-2x+3

The answer is 1/2x +5 but I dont know how they got that.

2nd question

Through (6,-6); parallel to 5x+6y=18

the answer is -5/6x-1

I got y=-5/6x-1

My teacher gave me the cheat sheet for the review to make sure when I study that I can correct them.

I'll show you how to do the first, you can use that to answer the second.

The first function is f(x) = -2x + 3, so the gradient of that line is -2.

The product of the gradients of perpendicular lines is -1, so that means the gradient of the normal must be 1/2 (since 1/2 x -2 = -1).

So for your normal, so far you have y = (1/2)x + c.

You are given a point that lies on this normal, namely (x, y) = (-2, 4), so substituting this in gives

4 = (1/2) x (-2) + c
4 = -1 + c
5 = c.

Therefore the equation of the normal is y = (1/2)x + 5.

In function notation this is \displaystyle \begin{align*} \mathbf{R} \to \mathbf{R} \left\{ (x, y) | y = \frac{1}{2}x + 5 \right\} \end{align*}

Now for your second, start by getting your original line in the form y = mx + c.
• Dec 8th 2011, 04:53 PM
daword177
Re: Function notation
thank you !