Doing some math homework and not sure if I got some questions right, just want to double check my answers:
Find the perimeter of an isosceles triangle whose base is on the line y= -4 and whose vertex is the same as the vertex of the parabola f(x) = -3x^2+12x+11 if the end points of the base of the triangle are on the parabola.
No need to type in all the work as I just want to double check my final answer.
Thanks for your time!
Find the perimeter of an isosceles triangle whose base is on the line y= -4 and whose vertex is the same as the vertex of the parabola f(x) = -3x^2+12x+11 if the end points of the base of the triangle are on the parabola.
No need to type in all the work as I just want to double check my final answer.
Thanks for your time!
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I get for vertices: ((-1, -4), (5, -4), (2, 23)
So our base is 6, two iso sides are: h2 = 27^2 + 9 = √738
P = 2√738 + 6
So our base is 6, two iso sides are: h2 = 27^2 + 9 = √738
P = 2√738 + 6
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f(x) = -3x^2+12x+11
x = -b/2a
= -12/-6 = 2
f(2) = 23
vertex at:(2 , 23)
-3x^2 + 12x + 11 = -4
x^2 - 4x - 5 = 0
(x + 1)(x - 5) = 0
x = -1 , 5
hence triangle ABC verteces coordinates are:
A(2 , 23) , B(-1 , -4) , C(5 , -4)
AB = AC = √ [(-1 - 2)² + (-4 - 23)²]
= 3√82
BC = |-1-5| = 6
P = 2* 3√82 + 6
= 6(√82 + 1)
≈ 60.33
x = -b/2a
= -12/-6 = 2
f(2) = 23
vertex at:(2 , 23)
-3x^2 + 12x + 11 = -4
x^2 - 4x - 5 = 0
(x + 1)(x - 5) = 0
x = -1 , 5
hence triangle ABC verteces coordinates are:
A(2 , 23) , B(-1 , -4) , C(5 , -4)
AB = AC = √ [(-1 - 2)² + (-4 - 23)²]
= 3√82
BC = |-1-5| = 6
P = 2* 3√82 + 6
= 6(√82 + 1)
≈ 60.33