Use the vertex and intercepts to determine the graph of the quadratic function. Give the equation for the parabola's axis of symmetry. Use the parabola to identify the function's domain and range.
f(x) = 4(x - 2)^2 - 4
a. The vertex of the parabola is?
b. the x-intercept(s) is/are?
c. The y-intercept of the parabola is?
d. The axis of symmetry is?
e. The domain of f is? (answer in interval notation)
f. The range of f is? (answer in interval notation)
need help remembering this stuff trying to get into a university.
f(x) = 4(x - 2)^2 - 4
a. The vertex of the parabola is?
b. the x-intercept(s) is/are?
c. The y-intercept of the parabola is?
d. The axis of symmetry is?
e. The domain of f is? (answer in interval notation)
f. The range of f is? (answer in interval notation)
need help remembering this stuff trying to get into a university.
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Since your parabola is of the form
f(x) = a(x - h)² + k, it should be pretty easy to figure out part a.
The vertex occurs at (h, k).
--- --- For f(x) = 4(x - 2)² - 4, the vertex is at (2, -4).
To find the x-intercepts, set y = f(x) = 0, and solve for x:
y = 4(x - 2)² - 4
0 = 4(x - 2)² - 4
4 = 4(x - 2)²
1 = (x - 2)²
±√1 = x - 2
±1 = x - 2
x = 3 and x = 1
--- --- x-intercepts are (3, 0) and (1, 0).
As for the y-intercepts, the process is the same but in reverse. Set x = 0 and solve for y:
y = 4(x - 2)² - 4
y = 4(0 - 2)² - 4
y = 4(4) - 4
y = 16 - 4
y = 12
--- --- y-intercept is (0, 12).
The axis of symmetry is the invisible vertical line that cuts the parabola in half. It can be identified by simply looking at the x component of the vertex, which is 2.
--- --- Axis of symmetry is x = 2.
A parabola is defined for all real values of x.
--- --- Domain = all real numbers, or (-∞, ∞).
Based on the general shape of a parabola, its vertex is either the minimum or the maximum value of the function. The "direction" of the parabola, meaning whether it opens up or down, is determined by the sign of a; negative indicates downward, positive indicates upward. (If you're past precalc, you can determine this using the derivative tests.) Since a is 4 (positive), the direction is upward, meaning that the vertex is the minimum point of the function.
The parabola's range of values for f(x) is therefore all values of y greater than and including -4, since that's the y component of the vertex.
--- --- Range = y ≥ -4, or [-4, ∞).
f(x) = a(x - h)² + k, it should be pretty easy to figure out part a.
The vertex occurs at (h, k).
--- --- For f(x) = 4(x - 2)² - 4, the vertex is at (2, -4).
To find the x-intercepts, set y = f(x) = 0, and solve for x:
y = 4(x - 2)² - 4
0 = 4(x - 2)² - 4
4 = 4(x - 2)²
1 = (x - 2)²
±√1 = x - 2
±1 = x - 2
x = 3 and x = 1
--- --- x-intercepts are (3, 0) and (1, 0).
As for the y-intercepts, the process is the same but in reverse. Set x = 0 and solve for y:
y = 4(x - 2)² - 4
y = 4(0 - 2)² - 4
y = 4(4) - 4
y = 16 - 4
y = 12
--- --- y-intercept is (0, 12).
The axis of symmetry is the invisible vertical line that cuts the parabola in half. It can be identified by simply looking at the x component of the vertex, which is 2.
--- --- Axis of symmetry is x = 2.
A parabola is defined for all real values of x.
--- --- Domain = all real numbers, or (-∞, ∞).
Based on the general shape of a parabola, its vertex is either the minimum or the maximum value of the function. The "direction" of the parabola, meaning whether it opens up or down, is determined by the sign of a; negative indicates downward, positive indicates upward. (If you're past precalc, you can determine this using the derivative tests.) Since a is 4 (positive), the direction is upward, meaning that the vertex is the minimum point of the function.
The parabola's range of values for f(x) is therefore all values of y greater than and including -4, since that's the y component of the vertex.
--- --- Range = y ≥ -4, or [-4, ∞).