Q, 14 - Precalculus Enhanced with Graphing Utilities

Question

(a) Find a linear function that contains the points (-2, 3)

and (1, -6). What is the slope? What are the intercepts

of the function? Graph the function. Be sure to label the

intercepts.

(b) Find a quadratic function that contains the point (-2, 3)

with vertex (1, -6). What are the intercepts of the

function? Graph the function.

(c) Show that there is no exponential function of the form

f(x) = aex that contains the points (-2, 3) and (1, -6).

Step-by-Step Solution

Answer

part(a) (1)The linear function is :$3x+y=-3$

(2)slope of the function is -3

(3) Intercepts points$\left(x,y\right)=\left(-1,-3\right)$

(4) 

part(b) (1) The quadratic function is :$y={\left(x-1\right)}^2-6$

(2) the intercepts points are$\left(x,y\right)=\left(\pm \sqrt{6}+1,-5\right)$

(3) 

Part(a)Step 1. Finding the linear equation from two points

we know the equation of a line passing through two points is

$y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)x-x_1$ 

so from question 

equation of the line is

$$\begin{gathered} y-3 = \left(\frac{-6-3}{1-\left(-2\right)}\right)x-\left(-2\right) \\ y-3=\left(-3\right)\left(x+2\right) \\ 3x+y =-3 \end{gathered}$$

for the slope of the curve, compare the equation with $y=mx$

so, the slope of the line is - 3 

part(a) step2. Intercepts points

or the intersection point 

for x intersects point

for y intersects point 

part(a) Step3. Graph of function i

part(b)step 1.Given information

point (-2,3)

vertex point (1,-6)

let's assume the  equation of quadratic function is$y={(x-a)}^2+b^2$

vertex of standard parabolic function( $x^2=4ay$ ) is (0,0)

compare  with the standard form

$$\begin{gathered} x-a = 0 \\ x= a \\ y-b=0 \\ y=b \end{gathered}$$

so vertex is (a,b)

but in question vertex is given (1, -6)

so quadratic equation is

 $y ={\left(x-a\right)}^2-6$

part(b)step 2. Calculating intercept points

the x-intersects point is 

for y intersects point

part(b) step3.

the graph of this function is  

part(c) step 1. Assuming the exponential function

let's assume the exponential function

the function contains the point (-2,3)

so 

similar 

the function contains the point (1,-6)

so

from the first equation, we find that

so the product of A and b cant be negative 

so therefore no exponential function like  contains these points