Answer:
5) DE = 7 units and DF = 4 units
6) ST = 8 units
![\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units](https://tex.z-dn.net/?f=%5Ctextsf%7B7%29%7D%20%5Cquad%20%5Csf%20OM%3D%5Cdfrac%7B3%7D%7B2%7D%5C%3Aunits)
8) x ≤ -3 and x ≥ 3
Step-by-step explanation:
<u>Information from Parts 1-4:</u>
brainly.com/question/28193969
- A = (3, 0) and C = (-3, 0)
<h3><u>Part (5)</u></h3>
Points A and D are the <u>points of intersection</u> of the two functions.
To find the x-values of the points of intersection, equate the two functions and solve for x:
![\implies g(x)=f(x)](https://tex.z-dn.net/?f=%5Cimplies%20g%28x%29%3Df%28x%29)
![\implies x^2-9=-x+3](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2-9%3D-x%2B3)
![\implies x^2+x-12=0](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2%2Bx-12%3D0)
![\implies x^2+4x-3x-12=0](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2%2B4x-3x-12%3D0)
![\implies x(x+4)-3(x+4)=0](https://tex.z-dn.net/?f=%5Cimplies%20x%28x%2B4%29-3%28x%2B4%29%3D0)
![\implies (x-3)(x+4)=0](https://tex.z-dn.net/?f=%5Cimplies%20%28x-3%29%28x%2B4%29%3D0)
Apply the zero-product property:
![\implies (x-3)= \implies x=3](https://tex.z-dn.net/?f=%5Cimplies%20%28x-3%29%3D%20%5Cimplies%20x%3D3)
![\implies (x+4)=0 \implies x=-4](https://tex.z-dn.net/?f=%5Cimplies%20%28x%2B4%29%3D0%20%5Cimplies%20x%3D-4)
From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.
To find the y-value of point D, substitute the found value of x into one of the functions:
![\implies f(-4)=-(-4)=7](https://tex.z-dn.net/?f=%5Cimplies%20f%28-4%29%3D-%28-4%29%3D7)
Therefore, D = (-4, 7).
The length of DE is the difference between the y-value of D and the x-axis:
⇒ DE = 7 units
The length of DF is the difference between the x-value of D and the x-axis:
⇒ DF = 4 units
<h3><u>Part (6)</u></h3>
To find point S, substitute the x-value of point T into function g(x):
![\implies g(4)=(4)^2-9=7](https://tex.z-dn.net/?f=%5Cimplies%20g%284%29%3D%284%29%5E2-9%3D7)
Therefore, S = (4, 7).
The length ST is the difference between the y-values of points S and T:
![\implies ST=y_S-y_T=7-(-1)=8](https://tex.z-dn.net/?f=%5Cimplies%20ST%3Dy_S-y_T%3D7-%28-1%29%3D8)
Therefore, ST = 8 units.
<h3><u>
Part (7)</u></h3>
The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:
![\implies f(x)-g(x)=QR](https://tex.z-dn.net/?f=%5Cimplies%20f%28x%29-g%28x%29%3DQR)
![\implies -x+3-(x^2-9)=\dfrac{45}{4}](https://tex.z-dn.net/?f=%5Cimplies%20-x%2B3-%28x%5E2-9%29%3D%5Cdfrac%7B45%7D%7B4%7D)
![\implies -x+3-x^2+9=\dfrac{45}{4}](https://tex.z-dn.net/?f=%5Cimplies%20-x%2B3-x%5E2%2B9%3D%5Cdfrac%7B45%7D%7B4%7D)
![\implies -x^2-x+\dfrac{3}{4}=0](https://tex.z-dn.net/?f=%5Cimplies%20-x%5E2-x%2B%5Cdfrac%7B3%7D%7B4%7D%3D0)
![\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)](https://tex.z-dn.net/?f=%5Cimplies%20-4%5Cleft%28-x%5E2-x%2B%5Cdfrac%7B3%7D%7B4%7D%5Cright%29%3D-4%280%29)
![\implies 4x^2+4x-3=0](https://tex.z-dn.net/?f=%5Cimplies%204x%5E2%2B4x-3%3D0)
![\implies 4x^2+6x-2x-3=0](https://tex.z-dn.net/?f=%5Cimplies%204x%5E2%2B6x-2x-3%3D0)
![\implies 2x(2x+3)-1(2x+3)=0](https://tex.z-dn.net/?f=%5Cimplies%202x%282x%2B3%29-1%282x%2B3%29%3D0)
![\implies (2x-1)(2x+3)=0](https://tex.z-dn.net/?f=%5Cimplies%20%282x-1%29%282x%2B3%29%3D0)
Apply the zero-product property:
![\implies (2x-1)=0 \implies x=\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cimplies%20%282x-1%29%3D0%20%5Cimplies%20x%3D%5Cdfrac%7B1%7D%7B2%7D)
![\implies (2x+3)=0 \implies x=-\dfrac{3}{2}](https://tex.z-dn.net/?f=%5Cimplies%20%282x%2B3%29%3D0%20%5Cimplies%20x%3D-%5Cdfrac%7B3%7D%7B2%7D)
As the x-value of points M, Q and P is negative, x = -³/₂.
Length OM is the difference between the x-values of points M and the origin O:
![\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}](https://tex.z-dn.net/?f=%5Cimplies%20x_O-x_m%3Do-%28-%5Cfrac%7B3%7D%7B2%7D%29%3D%5Cdfrac%7B3%7D%7B2%7D)
Therefore, OM = ³/₂ units.
<h3><u>Part (8)</u></h3>
The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.
Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.