Question

Can someone help me with 19-24... I need it to be worked out and answered... Please and thank you I really don't understand it, But it's due tomorrow morning so can someone please help... Thanks

Answer 1

Let me help you to understand this problem. Linear functions are those whose graph is a straight line. The form of a linear function is given by the following form:

$y=f(x)=ax+b \\ \\ where \ a \ and \ b \ are \ constants \\ \\ \\ a:Represents \ the \ slope \ of \ the \ line \\ \\ b:Represents \ the \ y-intercept$

The y-intercept occurs when x = 0.

So, our problem is to find the constants $a \ and \ b$ in each case. Therefore:

FIRST. (19)

$f(0)=2, \ f(2)=4$

The slope of a line can be found as follows:

$a=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} \\ \\ \\ f(x_{2})=f(2)=4 \\ \\ f(x_{1})=f(0)=2 \\ \\ \\ \therefore a=\frac{4-2}{2-0}=1$

As the y-intercept occurs when x = 0, therefore:

$b=f(0)=2$

Accordingly, the first linear function is:

$\boxed{y=f(x)=x+2}$

SECOND. (20)

$f(0)=7, \ f(3)=1$

Applying the same reasoning as in the first exercise, the slope is:

$f(x_{2})=f(3)=1 \\ \\ f(x_{1})=f(0)=7 \\ \\ \\ \therefore a=\frac{1-7}{3-0}=-2$

And the y-intercept is:

$b=f(0)=7$

Accordingly, the second linear function is:

$\boxed{y=f(x)=-2x+7}$

THIRD. (21)

$f(4)=-3, \ f(0)=-2$

The slope is:

$f(x_{2})=f(4)=-3 \\ \\ f(x_{1})=f(0)=-2 \\ \\ \\ \therefore a=\frac{-3-(-2)}{4-0}=-\frac{1}{4}$

And the y-intercept is:

$b=f(0)=-2$

Accordingly, the third linear function is:

$\boxed{y=f(x)=-\frac{1}{4}x-2}$

FOURTH. (22)

$f(5)=-1, \ f(0)=-5$

The slope is:

$f(x_{2})=f(5)=-1 \\ \\ f(x_{1})=f(0)=-5 \\ \\ \\ \therefore a=\frac{-1-(-5)}{5-0}=\frac{4}{5}$

And the y-intercept is:

$b=f(0)=-5$

Accordingly, the fourth linear function is:

$\boxed{y=f(x)=\frac{4}{5}x-5}$

FIFTH. (23)

$f(-2)=6, \ f(0)=-4$

The slope is:

$f(x_{2})=f(-2)=6 \\ \\ f(x_{1})=f(0)=-4 \\ \\ \\ \therefore a=\frac{6-(-4)}{-2-0}=-5$

And the y-intercept is:

$b=f(0)=-4$

Accordingly, the fifth linear function is:

$\boxed{y=f(x)=-5x-4}$

SIXTH. (24)

$f(0)=3, \ f(-6)=3$

The slope is:

$f(x_{2})=f(-6)=3 \\ \\ f(x_{1})=f(0)=3 \\ \\ \\ \therefore a=\frac{3-3}{-6-0}=0$

And the y-intercept is:

$b=f(0)=3$

Accordingly, the sixth linear function is:

$\boxed{y=f(x)=3} \\ \\ This \ is \ a \ constant \ function$