2. Let f(x)=ax^2+bx+c. What is the coordinate of the y-intercept of f(x)?
A. (0, -c )
B. (-b/2a,c)
C. (0,c)
D. (-b/2a,-c)
3. What is the vertex and x-intercept(s) of the function f(x)=x^2−2x−3?
A. The vertex is at (−1,−4) and the x-intercept is at (0,−3) .
B. The vertex is at (1,−4) and the x-intercepts are at (−1,0) and (3, 0)
C. The vertex is at (1,−4) and the x-intercept is at (0,−3) .
D. The vertex is at (−1,−4) and the x-intercepts are at (−1,0) and (3, 0)
2. The coordinate of the y-intercept of f(x) is (0, c).
3. The vertex is at (1,-4) and the x-intercepts are at (-1,0) and (3,0).
2. The coordinate of the y-intercept of f(x) is (0, c). Therefore, the correct answer is C. (0, c).
3. To find the vertex and x-intercepts of the function f(x) = x^2 - 2x - 3, we can start by completing the square:
f(x) = (x - 1)^2 - 1 - 3
= (x - 1)^2 - 4
The vertex form of a quadratic function is given by f(x) = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola. Comparing the given function f(x) to the vertex form, we can see that the vertex is located at (1, -4).
To find the x-intercepts, we set f(x) = 0 and solve for x:
0 = (x - 1)^2 - 4
4 = (x - 1)^2
2 = x - 1 or -2 = x - 1
x = 3 or x = -1
Therefore, the vertex is at (1, -4) and the x-intercepts are at (-1, 0) and (3, 0). The correct answer is D. The vertex is at (−1,−4) and the x-intercepts are at (−1,0) and (3, 0).
To find the coordinate of the y-intercept of a function, we need to find the value of the function when x = 0.
For question 2, we are given the function f(x) = ax^2 + bx + c. To find the y-intercept, we substitute x = 0 into the function.
f(0) = a(0)^2 + b(0) + c
f(0) = 0 + 0 + c
f(0) = c
Therefore, the coordinate of the y-intercept is (0, c).
Answer: C. (0, c)
Now, let's move on to question 3.
To find the vertex and x-intercepts of a quadratic function in the form f(x) = ax^2 + bx + c, we can use the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
Substituting this value into the function will give us the y-coordinate of the vertex.
For question 3, we are given the function f(x) = x^2 - 2x - 3.
To find the vertex, we use the formula:
x = -(-2) / (2*1)
x = 2 / 2
x = 1
Substituting x = 1 into the function:
f(1) = (1)^2 - 2(1) - 3
f(1) = 1 - 2 - 3
f(1) = -4
So the vertex is (1, -4).
To find the x-intercepts, we set f(x) = 0 and solve for x.
x^2 - 2x - 3 = 0
This equation can be factored as:
(x - 3)(x + 1) = 0
Setting each factor equal to 0, we have:
x - 3 = 0 -> x = 3
x + 1 = 0 -> x = -1
So the x-intercepts are (3, 0) and (-1, 0).
Answer: D. The vertex is at (-1, -4) and the x-intercepts are at (-1, 0) and (3, 0).