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.
(Bonus:intersection with f(x) =-2x²+4)
A) To find the zeros of the function, we set f(x) equal to zero and solve for x:
4x² + 16x - 12 = 0
Using the quadratic formula or factoring, we find the zeros to be x = -2 and x = 1.5.
B) The vertex of the quadratic function occurs at the x-coordinate given by the formula x = -b/2a. In this case, a = 4 and b = 16, so the x-coordinate of the vertex is x = -16/(2*4) = -2. The corresponding y-coordinate can be found by substituting this x-value into the function:
f(-2) = 4(-2)² + 16(-2) - 12 = -44
Therefore, the vertex is (-2, -44).
C) To write the equation in vertex form, we have to complete the square. We start with the function in standard form:
f(x) = 4x² + 16x - 12
Now, we can factor out the common factor of 4 from the first two terms:
f(x) = 4(x² + 4x) - 12
Completing the square on the quadratic term within the parentheses, we add and subtract the square of half the coefficient of x (in this case, 4/2 = 2) squared:
f(x) = 4(x² + 4x + 2² - 2²) - 12
Next, we can simplify and factor the terms inside the parentheses:
f(x) = 4((x + 2)² - 4) - 12
Expanding and combining like terms, we get:
f(x) = 4(x + 2)² - 16 - 12
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 a quadratic function is all real numbers, so the domain for this function is (-∞, ∞). The range can be found by considering the vertex (which gives the minimum or maximum point of the function depending on the shape of the parabola). Since the coefficient of the x² term is positive, the parabola opens upward, thus the minimum value occurs at the vertex. Therefore, the range is (-∞, -44].
E) The transformations on f(x) can be determined from the vertex form of the function: f(x) = 4(x + 2)² - 28. The vertex form reveals that the parabola has been shifted horizontally 2 units to the left (compared to the standard form equation) and shifted downward 28 units. The coefficient 4 in front of the squared term indicates that the graph has been vertically stretched by a factor of 4.
F) To find the intersection between f(x) = 4x² + 16x - 12 and f(x) = 1.5x + 16, we set the two equations equal to each other and solve for x:
4x² + 16x - 12 = 1.5x + 16
Subtracting 1.5x and 16 from both sides:
4x² + 14.5x - 28 = 0
Using the quadratic formula or factoring, we find the two x-values that represent the intersection points.