Determine the vertex,axis symmetry, x-intercept, and y-intercept for the equation. y = x^2 + 2x - 3
To determine the vertex of the equation y = x^2 + 2x - 3, we can first find the x-coordinate of the vertex by using the formula x = -b / (2a), where a, b, and c are the coefficients of the equation in standard form (ax^2 + bx + c = 0).
In this case, a = 1 and b = 2, so the x-coordinate of the vertex is x = -2 / (2 * 1) = -1.
Next, we substitute this x-coordinate into the equation to find the y-coordinate of the vertex: y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4.
Therefore, the vertex of the equation is (-1, -4).
The axis of symmetry is a vertical line that passes through the vertex. In this case, the equation is x = -1.
To find the x-intercepts, we set y equal to zero and solve the equation for x:
0 = x^2 + 2x - 3
This equation can be factored as (x - 1)(x + 3) = 0.
Setting each factor equal to zero, we find x - 1 = 0 or x + 3 = 0.
Solving these equations, we get x = 1 or x = -3.
Therefore, the x-intercepts of the equation are x = 1 and x = -3.
To find the y-intercept, we set x equal to zero in the equation:
y = (0)^2 + 2(0) - 3 = 0 + 0 - 3 = -3.
Therefore, the y-intercept of the equation is y = -3.
Summary:
Vertex: (-1, -4)
Axis of Symmetry: x = -1
x-intercepts: x = 1 and x = -3
y-intercept: y = -3