Which linear function includes the points (–3, 1) and (–2, 4)?

slope = [ (4 - 1) / ((- 2) - (- 3)) ] 3 / 1 = 3

⇔

y = 3x + k

Through point (- 3 ; 1) :

y = 3x + k [ with k = constant = y-intercept ]

1 = 3 (- 3) + k

1 = - 9 + k

k = 10

Answer :

y = f (x) = 3x + 10

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Use the formula of y = mx + b, where m is the slope and b is the y-intercept.

To find the slope, subtract the y values and divide by the subtraction of the x values:

slope = m = (4 - 1) / (-2 - -3) = 3/1 = 3

Now you have your slope.

But in your equation now, y = 3x + b, you need to solve for b. You can do this by plugging in values for x and y, let's use (-3, 1)

1 = 3*(-3) + b

b = 10

You are left with y = 3x + 10, or f(x) = 3x + 10

Two ways to solve this problem. One is the "brute force" method. Plug the given values into each equation and see which one is solved by both values.

Another way is the use the points to derive the equation of the line and see which of the given functions matches that equation.

I'd suggest using the two points to solve for m and b in the slope,intercept equation of a line

y = f(x) = mx + b

m = (y1-y0)/(x1-x0)

b = y0 - mx0

Once you know m and b, compare those values to the equivalent values in the given functions and see which one matches.