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

Dec 12, 2016

 "y=3x+10

#### Explanation:

Linear => straight line graph type function:

$\text{ } \to y = m x + c$ .................Equation(1)

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 3 , 1\right)$

Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(- 2 , 4\right)$

Substitute both these ordered pairs into equation(1) giving two new equation.
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$\textcolor{b l u e}{\text{Determine the gradient } m}$

${P}_{1} \to 1 = m \left(- 3\right) + c$ ................................Equation(2)
${P}_{2} \to 4 = m \left(- 2\right) + c$.................................Equation(3)

$E q u a t i o n \left(3\right) - E q u a t i o n \left(2\right)$

$4 - 1 = - 2 m + 3 m$

$\textcolor{b l u e}{3 = m \to m = 3}$

$\textcolor{b r o w n}{\text{Check: - alternative method}}$

Gradient $\to m = \left(\text{change in up or down")/("change in along}\right)$

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{4 - 1}{- 2 - \left(- 3\right)} = \frac{3}{1} = 3$
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$\textcolor{b l u e}{\text{Determine the constant (y-intercept) } c}$

${P}_{2} \to 4 = m \left(- 2\right) + c$

But $m = 3$

$\implies 4 = \left(3\right) \left(- 2\right) + c$

$4 = - 6 + c$

$\textcolor{b l u e}{c = 10}$
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$\textcolor{b l u e}{\text{Putting it all together}}$

$\textcolor{b l u e}{y = m x + c \text{ "->" } y = 3 x + 10}$