For the function f(x)=4x²+16x-12
A) What are the zeros?
B) what is the vertex?
C) write in vertex form
D) state the domain and range
E) what are the transformations on f(x)
F) find the intersection of the above function and f(x)=1.5x+16 algebraically
A) To find the zeros, set f(x) equal to zero and solve for x:
4x² + 16x - 12 = 0
You can factor the equation:
4(x² + 4x - 3) = 0
4(x + 3)(x - 1) = 0
Setting each factor equal to zero gives you:
x + 3 = 0, x = -3
x - 1 = 0, x = 1
Therefore, the zeros of the function f(x) = 4x² + 16x - 12 are -3 and 1.
B) The vertex of the function can be found using the formula for the x-coordinate of the vertex:
x = -b/2a
For f(x) = 4x² + 16x - 12, a = 4 and b = 16. Plugging in these values:
x = -(16)/(2*4)
x = -16/8
x = -2
To find the y-coordinate of the vertex, substitute this x-value back into the original equation:
f(-2) = 4(-2)² + 16(-2) - 12
f(-2) = 4(4) - 32 - 12
f(-2) = 16 - 32 - 12
f(-2) = -28
Therefore, the vertex of the function f(x) = 4x² + 16x - 12 is (-2, -28).
C) To write the function in vertex form, we use the formula:
f(x) = a(x - h)² + k, where (h, k) is the vertex.
Substituting the vertex values we found in part B, we get:
f(x) = 4(x + 2)² - 28
Therefore, the function in vertex form is f(x) = 4(x + 2)² - 28.
D) The domain of the function is all real numbers, since there are no restrictions or excluded values. This can be written as (-∞, ∞).
The range can be determined by considering the graph of the function. Since the coefficient of the x² term is positive, the parabola opens upward, and the minimum y-value is at the vertex, -28. Therefore, the range of the function is (-28, ∞).
E) The transformations on f(x) can be determined by comparing it to the parent function f(x) = x²:
The factor of 4 in front of x² stretches the parabola vertically by a factor of 4. This makes the parabola narrower and steeper.
The +16x term shifts the parabola horizontally to the left by 16 units. The -12 term shifts the parabola vertically downward by 12 units.
F) To find the intersection of f(x) = 4x² + 16x - 12 and f(x) = 1.5x + 16, set these two equations equal to each other and solve for x algebraically:
4x² + 16x - 12 = 1.5x + 16
Rearranging the equation and combining like terms:
4x² + 14.5x - 28 = 0
This equation cannot be factored easily, so we can use the quadratic formula:
x = (-b ± √(b² - 4ac))/(2a)
For this equation, a = 4, b = 14.5, and c = -28. Plugging in these values:
x = (-14.5 ± √(14.5² - 4(4)(-28)))/(2(4))
Simplifying:
x = (-14.5 ± √(210.25 + 448))/(8)
x = (-14.5 ± √658.25)/(8)
Using a calculator or estimating the square root, we get:
x ≈ -4.12 or x ≈ 1.62
Therefore, the intersection of the function f(x) = 4x² + 16x - 12 and f(x) = 1.5x + 16 is approximately x = -4.12 and x = 1.62.