Determine the zeros of the quadratic function y=x^2+15x+36 . (When listing the smaller zero, remember that negative values are always smaller than positive values.)
To find the zeros of the quadratic function, we set y equal to zero and solve for x:
−x^2 + 16x - 39 = 0
We can use factoring or the quadratic formula to solve for x.
Factoring:
(x - 3)(x - 13) = 0
Setting each factor equal to zero:
x - 3 = 0 or x - 13 = 0
x = 3 or x = 13
Therefore, the zeros of the quadratic function are x = 3 and x = 13. The smaller zero is x = 3.
To find the key points on the graph of y = x^2 - 16x + 48, we can determine the vertex, x-intercepts, and y-intercept.
First, let's find the vertex. The vertex of a quadratic function in the form y = ax^2 + bx + c can be found using the formula x = -b/(2a). In this case, a = 1 and b = -16. Plugging these values into the formula gives:
x = -(-16) / (2 * 1)
x = 16 / 2
x = 8
To find the y-coordinate of the vertex, we substitute x = 8 into the equation:
y = (8)^2 - 16(8) + 48
y = 64 - 128 + 48
y = -16
So, the vertex is (8, -16).
Next, let's find the x-intercepts. These are the points where the graph intersects the x-axis. To find them, we set y = 0 and solve for x.
0 = x^2 - 16x + 48
This quadratic equation does not factor easily, so we can use the quadratic formula to find the x-intercepts:
x = (-(-16) ± sqrt((-16)^2 - 4(1)(48))) / (2(1))
x = (16 ± sqrt(256 - 192)) / 2
x = (16 ± sqrt(64)) / 2
x = (16 ± 8) / 2
Simplifying further, we get:
x = 8 ± 4
So, the x-intercepts are x = 12 and x = 4.
Lastly, let's find the y-intercept. This is the point where the graph intersects the y-axis. To find it, we set x = 0 and solve for y:
y = (0)^2 - 16(0) + 48
y = 48
So, the y-intercept is (0, 48).
In summary, the key points on the graph of y = x^2 - 16x + 48 are:
- Vertex: (8, -16)
- X-intercepts: (12, 0) and (4, 0)
- Y-intercept: (0, 48)
We can find the zeros of the quadratic function by setting y equal to zero and solving for x.
x^2 + 15x + 36 = 0
We can use factoring or the quadratic formula to solve for x.
Factoring:
(x + 3)(x + 12) = 0
Setting each factor equal to zero:
x + 3 = 0 or x + 12 = 0
x = -3 or x = -12
Therefore, the zeros of the quadratic function are x = -3 and x = -12. The smaller zero is x = -12, as negative values are always smaller than positive values.
Determine the zeros of the quadratic function y=−x^2+16x−39 . (When listing the smaller zero, remember that negative values are always smaller than positive values.)
Determine the vertex and axis of symmetry of y=x^2−10x+3 .
To find the vertex and axis of symmetry of the quadratic function, we need to convert the given equation into vertex form, which is in the form y = a(x - h)^2 + k.
For the given equation y = x^2 - 10x + 3, we can complete the square to rewrite it into vertex form:
y = (x^2 - 10x) + 3
= (x^2 - 10x + 25) - 25 + 3
= (x - 5)^2 - 22
Comparing this with the vertex form, we can determine the vertex and axis of symmetry based on the values of h and k.
From the equation (x - 5)^2 + K (the vertex form), we can see that the vertex is at the point (h, k), where h = 5 and k = -22. So the vertex is (5, -22).
The axis of symmetry is a vertical line that passes through the vertex. In this case, since the x-coordinate of the vertex is 5, the equation of the axis of symmetry is x = 5.
What point is symmetrical to (−1,−3) for the graph of y=x^2+12x+8 ?
To find the point that is symmetrical to (-1, -3) for the graph of y = x^2 + 12x + 8, we need to find the equation of the axis of symmetry and then reflect the given point across that axis.
The axis of symmetry of a quadratic function in the form y = ax^2 + bx + c is given by the equation x = -b/2a.
In this case, the equation is y = x^2 + 12x + 8, so a = 1 and b = 12. Plugging these values into the equation for the axis of symmetry gives:
x = -12 / (2 * 1)
x = -12/2
x = -6
The axis of symmetry is x = -6.
To find the point symmetrical to (-1, -3), we need to reflect it across the axis of symmetry.
The distance between the given point (-1, -3) and the axis of symmetry x = -6 is the same as the distance between the unknown point and x = -6.
The x-coordinate of the unknown point should be on the other side of x = -6, so it will have a positive x value.
Using the difference in x-coordinates:
Difference in x = -1 - (-6) = 5
The x-coordinate of the unknown point will be 5 units to the right of x = -6:
x = -6 + 5
x = -1
Since the axis of symmetry is a vertical line, the y-coordinate of the unknown point will be the same as the y-coordinate of the given point:
y = -3
Therefore, the point that is symmetrical to (-1, -3) for the graph of y = x^2 + 12x + 8 is (-1, -3).
Question What are the key points on the graph of y=x^2−16x+48 ? Name the vertex, x-intercepts, and y-intercept.
To determine the zeros of the quadratic function, we need to find the values of x for which y is equal to zero.
The quadratic function is given as y = x^2 + 15x + 36.
To find the zeros, we set y equal to zero and solve for x:
0 = x^2 + 15x + 36
Now, we attempt to factorize the quadratic expression. However, in this case, it is not easily factorable. So, we can use the quadratic formula to find the zeros of the function.
The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 15, and c = 36.
Substituting these values into the quadratic formula, we get:
x = (-15 ± √(15^2 - 4(1)(36))) / (2)(1)
x = (-15 ± √(225 - 144)) / 2
x = (-15 ± √81) / 2
x = (-15 ± 9) / 2
Now, we have two possibilities:
1. x = (-15 + 9) / 2 = -6 / 2 = -3
2. x = (-15 - 9) / 2 = -24 / 2 = -12
Therefore, the zeros of the quadratic function y = x^2 + 15x + 36 are x = -3 and x = -12. The smaller zero is x = -12, according to the instruction that negative values are always smaller than positive values.