Question

section 5.2 - systems of linear equations in three variables and applications
question 11 of 11 (1 point) | question attempt: 1 of unlimited
(a) show that the points (-4, 9), (-2, -5), and (1, 4) are not collinear by finding the slope between (-4, 9) and (-2, -5), and the slope between (-2, -5) and (1, 4).
(b) find an equation of the form ( y = ax^2 + bx + c ) that defines the parabola through the points.
(c) use a graphing utility to verify that the graph of the equation in part (b) passes through the given points.
part: 0 / 3
part 1 of 3
(a) the slopes are and .

Explanation:

Step1: Calculate the slope between $(-4,9)$ and $(-2,-5)$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Here, $x_1=-4,y_1 = 9,x_2=-2,y_2=-5$.
$m_1=\frac{-5 - 9}{-2-(-4)}=\frac{-14}{2}=-7$.

Step2: Calculate the slope between $(-2,-5)$ and $(1,4)$

Using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Here, $x_1=-2,y_1=-5,x_2 = 1,y_2=4$.
$m_2=\frac{4-(-5)}{1-(-2)}=\frac{9}{3}=3$.

Answer:

The slopes are $-7$ and $3$.