Asked by mysterychicken
How many solutions does this have:
x^2 + 2x + 1 = 0
I put one.
9x^2 + 49 = 42x
I put one.
x^2 + 4x + 1 = 0
I put two.
x^2 + 2x + 1 = 0
I put one.
9x^2 + 49 = 42x
I put one.
x^2 + 4x + 1 = 0
I put two.
Answers
Answered by
mysterychicken
I actually think the 2nd one has no real solutions...
Answered by
Steve
nah, you are correct on all three
9x^2 - 42x + 49 = 0
(3x-7)^2 = 0
9x^2 - 42x + 49 = 0
(3x-7)^2 = 0
Answered by
mysterychicken
Ok cool! you think you can show me how to do these?
1. What is the largest output value of f(x) possible for the following quadratic function? f(x) = - x 2 -6 x + 15.
2. What is the absolute value of the smallest output value of f(x) possible for the following quadratic function? f(x) = x 2 – 2 x - 8
1. What is the largest output value of f(x) possible for the following quadratic function? f(x) = - x 2 -6 x + 15.
2. What is the absolute value of the smallest output value of f(x) possible for the following quadratic function? f(x) = x 2 – 2 x - 8
Answered by
Steve
as you know, the vertex of the parabola
y = ax^2 + bx + c
is at x = -b/2a
#1 the parabola opens down, so the vertex is a maximum
#2 has a minimum at the vertex
so, plug in that value for x and let 'er rip
y = ax^2 + bx + c
is at x = -b/2a
#1 the parabola opens down, so the vertex is a maximum
#2 has a minimum at the vertex
so, plug in that value for x and let 'er rip
Answered by
mysterychicken
I have no idea what any of that means... :)
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