y=x^2 - 2x - 3
y-intercept?
x-intercept?
axis of symmetry?
vertex?
If that's true, you are in deep doo doo
x^2-2x-3 = 0
(x-3)(x+1) = 0
x = -1, 3
-b/2a = 2/2 = 1
so the line x=1 is the axis of symmetry.
Note that it is midway between the x-intercepts
This is the vertex form of the equation.
Oops. It should be y = (x-1)^2 - 4
You can just read off the vertex.
It is at (1,-4)
Better get back to studying.
x-intercepts: find x such that x^2-2x-3 = 0
y-intercept: plug in x=0 and find y
axis of symmetry: x = -b/2a
vertex: note that x^2-2x-3 = (x^2-2x+1)-4 = (x-2)^2 - 4
That should get you going.
I don't have the brain capacity to get anything besides the y-intercept
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To find the y-intercept of a quadratic equation, you need to substitute x=0 into the equation and solve for y.
In this case, we have the equation y = x^2 - 2x - 3. To find the y-intercept, let's set x=0:
y = (0)^2 - 2(0) - 3
y = -3
Therefore, the y-intercept is -3.
To find the x-intercepts (also known as the roots or zeros) of a quadratic equation, you need to set y=0 and solve for x.
In this equation, y = x^2 - 2x - 3. Setting y=0:
0 = x^2 - 2x - 3
To solve this quadratic equation, you can factor it or use the quadratic formula. However, this equation cannot be easily factored, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -2, and c = -3. Substituting these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(1)(-3))) / (2(1))
x = (2 ± √(4 + 12)) / 2
x = (2 ± √16) / 2
x = (2 ± 4) / 2
We have two solutions:
x = (2 + 4) / 2 = 6 / 2 = 3
x = (2 - 4) / 2 = -2 / 2 = -1
Therefore, the x-intercepts are x = 3 and x = -1.
The axis of symmetry is a vertical line that passes through the vertex of the quadratic equation. The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
For our equation, a = 1 and b = -2. Substituting these values into the formula:
x = -(-2) / (2(1))
x = 2 / 2
x = 1
Therefore, the axis of symmetry is x = 1.
To find the y-coordinate of the vertex, you can substitute the x-coordinate into the equation:
y = (1)^2 - 2(1) - 3
y = 1 - 2 - 3
y = -4
Therefore, the vertex is located at (1, -4).
Help y=-x^2+2x-5
y=−x
2
+2x−5