1. The y-intercept of the function f(x) = -x^2 + 5x - 6 is when x = 0. Plugging in x = 0 into the function gives us f(0) = -0^2 + 5(0) - 6 = -6.
To find the point symmetric to the y-intercept, we reflect the y-intercept across the y-axis. The y-intercept is at (0, -6) and the point symmetric to it will be (-0, 6) which is the same as (0, 6).
The correct answer is C. (0, 6).
2. The y-intercept is the point where x = 0. When x = 0, the function f(x) = ax^2 + bx + c becomes f(0) = a(0)^2 + b(0) + c = c.
The coordinate of the y-intercept is therefore (0, c).
The correct answer is C. (0, c).
3. To find the vertex of the function f(x) = x^2 - 2x - 3, we need to find the x-coordinate of the vertex. The x-coordinate of the vertex is given by the formula x = -b/2a.
In this case, a = 1 and b = -2. Plugging these values into the formula gives x = -(-2)/(2*1) = 2/2 = 1.
To find the y-coordinate of the vertex, we plug the x-coordinate into the function f(x). f(1) = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4.
Therefore, the vertex is at (1, -4).
To find the x-intercepts, we set the function equal to 0 and solve for x. x^2 - 2x - 3 = 0.
Factoring or using the quadratic formula, we find that the x-intercepts are at x = -1 and x = 3.
The correct answer is D. The vertex is at (1, -4) and the x-intercepts are at (-1, 0) and (3, 0).
1. Where is the point symmetric to the y-intercept of the function f(x)=−x^2 +5x−6
A. (5/2,-6)
B. (-5,-6)
C. (5,-6)
D. (-5/2,-6)
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)
1 answer