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BECE Mathematics 2020 Paper

Question : 30

Total: 40

(-3,5)

(6,8)

Solution:

Equation of a straight line,

y=m x+c

where

m

is the gradient of the line

m=

change in

y /

change in

x=\frac{(y_2-y_1)}{(x_2-x_1)}

NOTE: The first coordinate represents the

x

-axis and the second the

y

-axis

Thus in

(-3,5), x=-3

and

y=5

\ln (6,8), x=6

and

y=8

\Rightarrow

The gradient of the line

=\frac{(8-5)}{(6--3)}=\frac{3}{(6+3)}=\frac{3}{9}=\frac{1}{3}

Use any of the points and the gradient to calculate the value of

c

Using the point

(-3,5)

\;\Rightarrow 5=\frac{1}{3}x-3+c

\;\Rightarrow 5=-1+c

\;\Rightarrow 5+1=c

\;\Rightarrow c=6

Using the point

(6,8)

\;\Rightarrow 8=\frac{1}{3} \times 6+c

\;\Rightarrow 8=2+c

\;\Rightarrow 8-2=c

\;\Rightarrow c=6

Hence the equation will be

y=\frac{1}{3^x}+6

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