30. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
30. f(-5) = -4, and f(5) = 2

Respuesta :

Answer:

The required linear equation satisfying the given conditions f(-5)=-4 and f(5)=2 is

[tex]$y=\frac{3}{5} x-1$[/tex]

Step-by-step explanation:

It is given that f(-5)=-4 and f(5)=2. It is required to find out a linear equation satisfying the conditions f(-5)=-4 and f(5)=2. To find it out, first, represent the given conditions in the form of points and then find the slope of a line passing through these two given points. Then consider one of the points to give the linear equation of the line in the form [tex]$\left(y-y_{2}\right)=m\left(x-x_{2}\right)$[/tex]

Step 1 of 4

Observe, f(-5)=-4 gives the point (5,-4)

And f(5)=2 gives the point (5,2)

This means that the function f(x) satisfies the points (-5,-4) and (5,2).

Step 2 of 4

Now find out the slope of a line passing through the points (-5,-4) and (5,2).

[tex]$\begin{aligned}m &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\m &=\frac{2-(-4)}{5-(-5)} \\m &=\frac{2+4}{5+5} \\m &=\frac{6}{10} \\m &=\frac{3}{5}\end{aligned}$[/tex]

Step 3 of 4

Now use the slope [tex]$m=\frac{3}{5}$[/tex] and use one of the two given points and write the equation in point-slope form:

[tex]$\left(y-y_{2}\right)=m\left(x-x_{2}\right)$\\ $(y-2)=\frac{3}{5}(x-5)$[/tex]

Distribute [tex]$\frac{3}{5}$[/tex],

[tex]$\begin{aligned}&y-2=\frac{3}{5} x-\frac{3}{5} \times 5 \\&y-2=\frac{3}{5} x-3\end{aligned}$[/tex]

Step 4 of 4

This linear function can be written in the slope-intercept form by adding 2 on both sides, [tex]$y-2+2=\frac{3}{5} x-3+2$[/tex]

[tex]$y=\frac{3}{5} x-1$[/tex]

So, this is the required linear equation.