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What function describes a line with slope 4/3 and passing through the point (2,5) ?

User Obchardon
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1 Answer

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For this case we have that by definition, the equation of the line of the slope-intersection form is given by:


y = mx + b

Where:

m: It's the slope

b: It is the cut-off point with the y axis

According to the data given we have:


m = \frac {4} {3}

Thus, the equation is of the form:


y = \frac {4} {3} x + b

We substitute the given point and find the cut point "b":


(x, y) :( 2,5)

So:


5 = \frac {4} {3} (2) + b\\5 = \frac {8} {3} + b\\5- \frac {8} {3} = b\\b = \frac {3 * 5-8} {3}\\b = \frac {15-8} {3}\\b = \frac {7} {3}

Thus, the equation is:


y = \frac {4} {3} x + \frac {7} {3}

Finally, we have that the function is of the form
y = f (x), where
f (x) = \frac {4} {3} x + \frac {7} {3}

ANswer:

The function that describes the line is:
f (x) = \frac {4} {3} x + \frac {7} {3}

User DFayet
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