Question

. Find the equation of the secant line on the graph of

Answer 1

Given:

`f(x)=3.5x^2-6x`

Line passes the point is:

`x_{1\text{ }}and_{}(\square)/(\square)x_2`
`\begin{gathered} x_1=5 \\ f(x_1)=3.5x^2_1-6x_1 \\ =3.5(5)^2-(6)5_{} \\ =87.5-30 \\ =57.5 \\ \text{ line pass=(5,57.5)} \end{gathered}`
`\begin{gathered} x_2=10 \\ f(x_2)=3.5(10)^2-6(10)_{} \\ =350-60 \\ =290 \\ \end{gathered}`

line is y = mx+c

`m=(y_2-y_1)/(x_2-x_1)`
`\begin{gathered} m=(290-57.5)/(10-5) \\ m=(232.5)/(5) \\ m=46.5 \end{gathered}`

Eq of line is:

y

`\begin{gathered} y=46.5x+c \\ 290=46.5*10+c \\ c=290-465 \\ c=-175 \end{gathered}`

So euation of line is:

`\begin{gathered} y=mx+c \\ y=46.5x-175 \end{gathered}`