What is an equation in point slope form of a line that passes through (-3,1) and has a slope of 2

y=2x+7

Step-by-step explanation:

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was 2. So you can right away fill in the equation for a line somewhat to read:

y=2x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(-3,1). When x of the line is -3, y of the line must be 1.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: . b is what we want, the 2 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (-3,1).

So, why not plug in for x the number -3 and for y the number 1? This will allow us to solve for b for the particular line that passes through the point you gave!.

(-3,1). y=mx+b or 1=2 × -3+b, or solving for b: b=1-(2)(-3). b=7.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!