Given triangle ABC with sides AB = 3x + 4 , BC = 2x + 5 , and AC = 4x , determine the values of x such that triangle ABC exists
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Answer: Answer is in the steps..
Step-by-step explanation:
f(x)= - 3x -5
A) To find the x intercepts of a parabola we have to set the whole equation equal zero because the x intercept is when y equal zero in other words where f(x) equal zero.
- 3x - 5 = 0 Now solve using quadratic formula.
x = -b±
x = 3 ±
x = 3 ± / 2a
x= 3 ± 7 /4
x= -1 or x = 5/2
The x intercepts are (5/2,0) and (-1,0)
B) The graph vertex of the graph of f(x) is going to be at a minimum because the leading coefficient of the function has a positive integer which means the parabola will open up and its vertex will be at the minimum point.
Using the x coordinates we could find their average and find the x coordinate of the vertex.
=
= 3/4
The x coordinate of the vertex is 3/4 so we will input that into the equation and solve for the y coordinate
= 18/16 - 9/4 -5 = -49/8
The y coordinate of the vertex is -49/8
So the vertex is (3/4, -49/8)
C) First I will graph the vertex and then graph he two x intercepts and connect then like a U shape.
