Question

For the quadratic relation y = -3x2 - 7x + 8,

a) state the y-intercept

b) determine the zeros, to 1 decimal place (show your process)

c) use your results from a) and b), to clearly graph and label this quadratic relation. Include your graph image or link here.

Answer 1

Answer:

a) y-int is at (0, 8)

b) zeros are at (0.8, 0) and (-3.2, 0); after having rounded to the nearest tenth.

Step-by-step explanation:

Given that y = -3x² - 7x + 8

we can find our y-intercept by setting x = 0

y = -3 (0)² - 7 (0) + 8

y = 8

so, our y intercept is at (8, 0)

To find our zeros, or x-intercepts, we need to set y = 0

0 = -3x² - 7x + 8

Let's use the quadratic formula

x = (-b ± √(b² - 4 (a  * c))) / 2a

where, in this case

a = -3

b = -7

c = 8

x = (7 ± √((-7)² - 4 (-3 * 8))) / (2 * -3)

x = (7 ± √(49 - -96) / -6

x = (7 ± √145) / -6

using the addition pathway

x = (7 + √145) / -6

x = 3.2

using the subtraction pathway

x = (7 - √145) / -6

x = -0.8

So, our x-intercepts, or zeros, will lie on the points

(0.8, 0) and (-3.2, 0)

Create a table of x and y values using the given equation, and plot and graph. Clearly label your y and x intercepts.

(x, y)

(-3, 2)

(-2, 10)

(-1, 12)

(0, 8)

(1, -2)

(2, -18)

(3, -40)

Answer 2

Answer:

y intercept  (0,8)

zero's ( -3.2 ,0) (.8,0)

Step-by-step explanation:

y = -3x2 - 7x + 8,

to find the y intercept, set x = 0

y = 0-0+8

the y intercept is 8

y = -3x2 - 7x + 8,

use the quadratic equation to find the zeros

-b ±sqrt(b^2 -4ac)

--------------------------

         2a

-(-7) ±sqrt((-7)^2 -4*(-3)*8)

---------------------------------

           2*(-3)

(7) ±sqrt((49 +96)

---------------------------------

           -6

(7) ±sqrt((145)

---------------------------------

           -6

(7+ sqrt(145))/-6  =-3.173599       round to 1 decimal place   -3.2

and

(7- sqrt(145))/ -6=.840266   round to 1 decimal place   .8