Complete parts a – c for each quadratic function.
a. Find the y-intercept, the equation of the axis of symmetry and the x-coordinate of the vertex.
b. Make a table of values that includes the vertex.
c. Use this information to graph the function.
1. f(x) = 2x2
2. f(x) = x2 + 4
3. f(x) = 2x2 – 4
4. f(x) = x2 – 4x + 4
5. f(x) = x2 – 4x – 5
6. f(x) = 3x2 + 6x – 1
7. f(x) = -3x2 – 4x
8. f(x) = 0.5x2 – 1
9. f(x) = ½x2 + 3x + 9/2
Determine whether each function has a maximum or a minimum value. Then find the maximum or minimum value of each function.
10. f(x) = 3x2
11. f(x) = x2 – 8x + 2
12. f(x) = 4x – x2 + 1
13. f(x) = 2x + 2x2 + 5
14. f(x) = -7 – 3x2 + 12x
15. f(x) = -½x2 – 2x + 3
I accidently posted the wrong thing, i will re-post the right assignment for help thankyou!!!
a. To find the y-intercept, we need to set x = 0 and solve for y. For example, for the function f(x) = 2x^2, when x = 0, we have f(0) = 2(0)^2 = 0. So the y-intercept is at (0, 0).
To find the equation of the axis of symmetry, we use the formula x = -b / (2a), where a and b are coefficients in the quadratic function. The equation of the axis of symmetry represents the vertical line that divides the parabola into two symmetrical halves. In this case, for f(x) = 2x^2, a = 2 and b = 0. Therefore, x = -0 / (2*2) = 0. The equation of the axis of symmetry is x = 0.
To find the x-coordinate of the vertex, we substitute the value of x from the equation of the axis of symmetry into the original function. For f(x) = 2x^2, when x = 0, we have f(0) = 2(0)^2 = 0. So the x-coordinate of the vertex is 0.
b. To make a table of values that includes the vertex, we can choose a few x-values around the vertex and calculate the corresponding y-values. For f(x) = 2x^2, we can use x = -1, 0, and 1.
For x = -1, f(-1) = 2(-1)^2 = 2.
For x = 0, f(0) = 2(0)^2 = 0.
For x = 1, f(1) = 2(1)^2 = 2.
So the table of values for f(x) = 2x^2 would be:
x | f(x)
---------
-1 | 2
0 | 0
1 | 2
c. To graph the function, we plot the points from the table of values on a coordinate plane. For f(x) = 2x^2, we plot the points (-1, 2), (0, 0), and (1, 2). We can also plot the y-intercept, (0, 0).
Connecting these points, we get a parabola that opens upwards. The vertex is at (0, 0) and the axis of symmetry is the vertical line x = 0. The graph would look like a U-shape opening upwards.
Now, let's go through the process for the other quadratic functions and determine whether each has a maximum or minimum value.
10. f(x) = 3x^2
- The y-intercept is (0, 0).
- The axis of symmetry is x = 0.
- The x-coordinate of the vertex is 0.
- This function has a minimum value at the vertex, which is (0, 0).
11. f(x) = x^2 - 8x + 2
- The y-intercept is (0, 2).
- To find the axis of symmetry, we use the formula x = -b / (2a). In this case, a = 1 and b = -8. Therefore, x = -(-8) / (2*1) = 4. The equation of the axis of symmetry is x = 4.
- To find the x-coordinate of the vertex, we substitute the value of x from the equation of the axis of symmetry into the original function. For x = 4, f(4) = (4)^2 - 8(4) + 2 = -14. So the x-coordinate of the vertex is 4.
- This function has a minimum value at the vertex, which is (4, -14).
12. f(x) = 4x - x^2 + 1
- The y-intercept is (0, 1).
- To find the axis of symmetry, we use the formula x = -b / (2a). In this case, a = -1 and b = 4. Therefore, x = -4 / (2*(-1)) = 2. The equation of the axis of symmetry is x = 2.
- To find the x-coordinate of the vertex, we substitute the value of x from the equation of the axis of symmetry into the original function. For x = 2, f(2) = 4(2) - (2)^2 + 1 = 5. So the x-coordinate of the vertex is 2.
- This function has a maximum value at the vertex, which is (2, 5).
(Continued in the next message)