Given: Line A R bisects ∠BAC; AB = AC Triangle A B C is shown. Point R is at the middle of the triangle. Lines are drawn from po
2 answers:
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Answer:
see explanation
Step-by-step explanation:
(a)
The zeros are x = 1 and x = 5, thus the factors are (x - 1) and (x - 5)
f(x) = a(x - 1)(x - 5)
To find a substitute (0, 15), the coordinates of the y- intercept into the equation.
15 = a(0 - 1)(0 - 5) = a(- 1)(- 5) = 5a ( divide both sides by 5)
a = 3
Thus h = 1, k = 5, a = 3 and
f(x) = 3(x - 1)(x - 5)
(b)
The axis of symmetry is a vertical line with equation x = c ( c is a constant )
The axis of symmetry passes through the midpoint of the zeros
x = =
= 3
Equation of axis of symmetry is x = 3
(c)
The axis of symmetry passes through P, with x- coordinate x = 3
Substitute x = 3 into the function for corresponding value of y
f(3) = 3(3 - 1)(3 - 5) = 3(2)(- 2) = - 12
Coordinates of P = (3, - 12 )