how do you find the x-intercepts y=x^2+2x-3
at the x-intercept, y = 0
so solve
x^2 + 2x - 3 = 0
The factors are obvious
I still am confused, but let me try this
(x-1)(x+2) but I am not sure where to go from here
lYou factored it correcty.
The only way (x-1)(x+2) can equal zero is for one OR the other of the two factors to be zero.
Therefore x = 1 or -2.
oh makes sense! Because once you put them on the opposite side of the equation they would change fromeither positive to negative or negative to positive.
I think the factors are
(x+3)(x-1)
so solutions are
-3 and +1
To find the x-intercepts of a quadratic function, you need to set the function equal to zero and solve for x. In this case, the given quadratic function is y = x^2 + 2x - 3.
Step 1: Set y equal to zero.
0 = x^2 + 2x - 3
Step 2: Factor the quadratic equation, if possible. In this case, the equation cannot be easily factored, so we'll use the quadratic formula.
Step 3: The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a.
For our given equation, a = 1, b = 2, and c = -3.
Step 4: Substitute the values of a, b, and c 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
Step 5: Simplify the expression.
x = (-2 + 4) / 2 or x = (-2 - 4) / 2
x = 2 / 2 or x = -6 / 2
x = 1 or x = -3
So, the x-intercepts of the given function y = x^2 + 2x - 3 are (1, 0) and (-3, 0).