Watch the video to go over the quadratic functions.
Determine the vertex, axis of symmetry, x-intercept, and y -intercept for the equation: `y=x^{2}+2x-3`.
To determine the vertex of a quadratic function in the form `y=ax^2+bx+c`, use the formula `x=-b/(2a)`.
In this case, `a=1` and `b=2`, so `x=-2/(2*1)= -1`.
To find the y-coordinate of the vertex, substitute the value of `x` into the equation:
`y=(-1)^2+2(-1)-3=1-2-3=-4`.
Therefore, the vertex is (-1,-4).
The axis of symmetry is the vertical line that passes through the vertex, which has the equation `x=-1`.
To find the x-intercepts, set `y` to zero and solve for x:
0=x^2+2x-3=(x+3)(x-1).
Setting each factor equal to zero, we get:
x+3=0 -> x=-3
x-1=0 -> x=3.
Therefore, the x-intercepts are (-3,0) and (1,0).
To find the y-intercept, substitute x=0 into the equation:
y=(0)^2+2(0)-3=0-3=-3.
Therefore, the y-intercept is (0,-3).