Question

If g(x) is a linear function such that g(-3) = 2 and g(1) = -4, find g(7).

Answer 1

Answer:   -13

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Explanation:

g(-3) = 2 means x = -3 and y = 2 pair up together to form the point (-3,2)

g(1) = -4 means we have the point (1,-4)

Find the slope of the line through the two points (-3,2) and (1,-4)

m = (y2-y1)/(x2-x1)

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

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

m = -6/4

m = -3/2

m = -1.5

The general slope intercept form y = mx+b turns into y = -1.5x+b after replacing m with -1.5

Plug in (x,y) = (-3,2) which is one of the points mentioned earlier and we end up with this new equation:  2 = -1.5*(-3) + b

Let's solve for b

2 = -1.5*(-3)+b

2 = 4.5 + b

2-4.5 = 4.5+b-4.5 .... subtract 4.5 from both sides

-2.5 = b

b = -2.5

Therefore, y = mx+b becomes y = -1.5x-2.5 meaning the g(x) function is g(x) = -1.5x-2.5

The last step is to plug in x = 7 and compute

g(x) = -1.5*x - 2.5

g(7) = -1.5*7 - 2.5

g(7) = -10.5 - 2.5

g(7) = -13

Answer 2

$\bf g(-3)=2\implies \begin{cases} x=-3\\ y=2 \end{cases}~\hspace{9em} g(1)=-4\implies \begin{cases} x=1\\ y=-4 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-2}{1-(-3)}\implies \cfrac{-6}{1+3}\implies -\cfrac{6}{4}\implies -\cfrac{3}{2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-\cfrac{3}{2}[x-(-3)] \\\\\\ y-2=-\cfrac{3}{2}(x+3)\implies y-2=-\cfrac{3}{2}x-\cfrac{9}{2}\implies y=-\cfrac{3}{2}x-\cfrac{9}{2}+2 \\\\\\ y=-\cfrac{3}{2}x-\cfrac{5}{2}$

$\bf \stackrel{y}{g(x)}=-\cfrac{3}{2}x-\cfrac{5}{2}\implies g(7)=-\cfrac{3}{2}(7)-\cfrac{5}{2}\implies g(7)=-\cfrac{21}{2}-\cfrac{5}{2} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill g(7)=-13~\hfill$