One x-intercept for a parabola is at the point (-2,0).
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Answer:
5) DE = 7 units and DF = 4 units
6) ST = 8 units
8) x ≤ -3 and x ≥ 3
Step-by-step explanation:
<u>Information from Parts 1-4:</u>
- A = (3, 0) and C = (-3, 0)
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:
Apply the zero-product property:
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:
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):
Therefore, S = (4, 7).
The length ST is the difference between the y-values of points S and T:
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:
Apply the zero-product property:
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:
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.