Which isa zero of the quadratic function f(x) = 4x + 24x + 11?
2 answers:
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The asymptotes are where the graph is undefined. Since: tan(x) =sin(x)/cos(x)
It is where cos(4x-π) = 0
cos(4x-π) = 0 when the inside is -π/2 , π/2 , 3π/2
4x - π = π/2
4x = π/2 + π
4x = 3π/2
x = 3π/8
4x - π = 3π/2
4x = 3π/2 + π
4x = 5π/2
x = 5π/8
This ones outside the interval (5π/8 > π/2) , try -π/2
4x - π = -π/2
4x = -π/2 + π
4x = π/2
x = π/8
Asymptotes are π/8 and 3π/8
Recall that the vertex form of a quadratic function (or parabolic function) is equal to
Now, given that we have f(x) = x² + 14x + 40, to express f into its vertex form, we must first fill in the expression to form a perfect square.
One concept that we must remember when completing the square is that
(a + b)² = a² + 2ab + b²
So, to complete the square for (x² + 14x + ____), we have 2ab = 14 where a = 1. Thus, b = 14/2 = 7. Hence, the last term of the perfect square must equal to 7² = 49.
So, going back to the function, we have
Thus, we have derived the vertex form of the function.
Answer: f(x) = (x + 7)² - 9
Now, given that we have f(x) = x² + 14x + 40, to express f into its vertex form, we must first fill in the expression to form a perfect square.
One concept that we must remember when completing the square is that
(a + b)² = a² + 2ab + b²
So, to complete the square for (x² + 14x + ____), we have 2ab = 14 where a = 1. Thus, b = 14/2 = 7. Hence, the last term of the perfect square must equal to 7² = 49.
So, going back to the function, we have
Thus, we have derived the vertex form of the function.
Answer: f(x) = (x + 7)² - 9