#1: f(x) = x^2-4x
find vrtex, axis ofsym, y-int, x-int if any, domain, range, where increasing and decreasing.
I got.... vertex: 2,4
axis of sym: x=2
D= (-infinity, infnity)
range= [4,infinity)
im stuck of the intercepts because when i plug i 0 for x i get yint of 0 but the graph doesnt show that.
Am i doing the completingthe square wrong
I got (x^2-2)^2 + 4
#2 Same requirements as the first question
f(x)= x^2 + 6x +9
I got... vertex: (-3,18)
axis of sym: x=-3
Domain= (-infinity, infinity)
Range= [18, infinity)
y-int (0,9)
x-int idk???
After completing the square I got:
(x+3)^2 + 18
Also a question, how do you know if there is an x or y int without drawing the graph because my graph des not show a xint but the equation does?
Thanks!
For question #1:
To find the vertex, axis of symmetry, y-intercept, x-intercepts (if any), domain, range, and where the function is increasing or decreasing, let's go step by step.
1. Vertex: To find the vertex of the quadratic function f(x) = x^2 - 4x, first identify the coefficient of x^2 (which is 1), and then use the formula: x = -b/2a. Plugging in the values, we get:
x = -(-4) / (2*1)
x = 4 / 2
x = 2
To find the y-coordinate of the vertex, substitute the value of x back into the original equation:
f(2) = (2)^2 - 4(2)
f(2) = 4 - 8
f(2) = -4
Therefore, the vertex is (2, -4).
2. Axis of symmetry: The axis of symmetry is the vertical line that passes through the vertex. In this case, the equation is x = 2.
3. Y-intercept: The y-intercept is the point at which the graph intersects the y-axis. To find it, set x = 0 in the equation:
f(0) = (0)^2 - 4(0)
f(0) = 0
So, the y-intercept is (0, 0).
4. X-intercept: To find the x-intercepts, set y = 0 in the equation and solve for x:
0 = x^2 - 4x
This equation can be factored as:
0 = x(x - 4)
So, the x-intercepts are x = 0 and x = 4.
5. Domain: The domain of the function is all real numbers, since there are no restrictions on the input x.
6. Range: From the vertex and the shape of the graph, we can see that the function is a upward-opening parabola, which means the minimum value occurs at the vertex. The range is all real numbers greater than or equal to the y-coordinate of the vertex. Therefore, the range is [ -4, infinity).
7. Increasing and decreasing: The function is decreasing for x less than 2 and increasing for x greater than 2.
Regarding your question about completing the square, to find the vertex, you can also use that method if it helps you. However, in this case, it seems like you might have made an error when completing the square, as the vertex you obtained is incorrect.
For question #2:
1. Vertex: Using the same steps as before, we have the equation f(x) = x^2 + 6x + 9.
First, find the x-coordinate of the vertex using x = -b/2a:
x = -6 / (2*1)
x = -3
Then, substitute x = -3 back into the equation to find the y-coordinate:
f(-3) = (-3)^2 + 6(-3) + 9
f(-3) = 9 - 18 + 9
f(-3) = 0
Therefore, the vertex is (-3, 0).
2. Axis of symmetry: The axis of symmetry is x = -3.
3. Y-intercept: Set x = 0 in the equation to find the y-intercept:
f(0) = (0)^2 + 6(0) + 9
f(0) = 9
So, the y-intercept is (0, 9).
4. X-intercept: Set y = 0 and solve for x:
0 = x^2 + 6x + 9
The equation can be factored as:
0 = (x+3)^2
So, the only x-intercept is x = -3.
5. Domain: The domain of the function is all real numbers, as there are no restrictions on the input x.
6. Range: Since the parabola opens upward and has its vertex at the lowest point, the y-coordinate of the vertex (0) represents the minimum value. Therefore, the range is [0, infinity).
7. Increasing and decreasing: The function is increasing for x less than -3 and decreasing for x greater than -3.
Regarding your question on how to determine the presence of x or y-intercepts without graphing, you can find the x-intercepts by setting y = 0 and solving the resulting equation. For the y-intercept, you set x = 0 and find the corresponding y-value. However, sometimes the graph may not accurately represent the exact location of the intercepts, especially when the function has complex roots or the scale of the graph is small. In those cases, it's better to rely on algebraic methods to find the intercepts accurately.
To find the vertex, axis of symmetry, and y-intercept of a quadratic function, you can use the vertex form of the equation, which is f(x) = a(x-h)^2 + k. In this form, (h, k) represents the vertex of the parabola. The axis of symmetry is the vertical line passing through the vertex, given by x = h. The y-intercept can be found by evaluating the function at x = 0, resulting in the point (0, f(0)).
Let's use this method to solve the first question, f(x) = x^2 - 4x:
Step 1: Find the vertex
To find the vertex, we need to complete the square. Rearrange the equation as follows:
f(x) = (x^2 - 4x + 4) - 4
= (x - 2)^2 - 4
From this form, we can see that the vertex is (2, -4).
Step 2: Find the axis of symmetry
The axis of symmetry is the vertical line passing through the vertex, given by x = 2.
Step 3: Find the y-intercept
Evaluate the function at x = 0:
f(0) = (0)^2 - 4(0)
= 0
Thus, the y-intercept is (0, 0).
Step 4: Find the x-intercept(s)
To find the x-intercepts, we set f(x) equal to zero and solve for x:
0 = (x - 2)^2 - 4
4 = (x - 2)^2
±2 = x - 2
Solving for x, we get x = 2 ± 2. This results in two x-intercepts: (0, 0) and (4, 0).
Step 5: Determine the domain
The domain of a quadratic function is all real numbers. So, the domain of f(x) is (-∞, ∞).
Step 6: Determine the range
Since the vertex has a y-coordinate of -4 and the parabola opens upward (since the leading coefficient a = 1 is positive), the range is [ -4, ∞).
Step 7: Determine where the function is increasing or decreasing
The function f(x) = x^2 - 4x is increasing to the right of the vertex. Therefore, it is increasing for x > 2. Similarly, it is decreasing for x < 2.
For the second question, f(x) = x^2 + 6x + 9:
Step 1: Find the vertex
Completing the square:
f(x) = (x^2 + 6x + 9) - 9
= (x + 3)^2 - 9
From this form, we can see that the vertex is (-3, -9).
Step 2: Find the axis of symmetry
The axis of symmetry is the vertical line passing through the vertex, given by x = -3.
Step 3: Find the y-intercept
Evaluate the function at x = 0:
f(0) = (0)^2 + 6(0) + 9
= 9
Thus, the y-intercept is (0, 9).
Step 4: Find the x-intercept(s)
Set f(x) equal to zero and solve:
0 = (x + 3)^2 - 9
9 = (x + 3)^2
±3 = x + 3
Solving for x, we get x = -3 ± 3. This results in two x-intercepts: (-6, 0) and (0, 0).
Step 5: Determine the domain
The domain of f(x) is (-∞, ∞).
Step 6: Determine the range
The vertex has a y-coordinate of -9, and since the parabola opens upward, the range is [ -9, ∞).
Step 7: Determine where the function is increasing or decreasing
The function f(x) = x^2 + 6x + 9 is increasing for x > -3 and decreasing for x < -3.
Regarding your question about determining the presence of x or y intercepts without graphing, the y-intercept is easily found by evaluating the function at x = 0. If the function results in y = 0, then there is a y-intercept at (0, 0). For x-intercepts, you need to set the function equal to zero and solve for x. If there are real solutions for x, then there are x-intercepts present. If the equation of the function suggests the presence of intercepts, but they don't show up on the graph, it may be due to limitations in scaling or the resolution of the graph.
I hope this explanation helps! Let me know if you have any further questions.