what would the x intercepts be for this equation?
y=x^2-3x-4
thanks.
set y to zero. (x-4)(x+1)=0 What are the roots?
To find the x-intercepts of the given equation, we first set y to zero and solve for x.
Given equation: y = x^2 - 3x - 4
Setting y to zero:
0 = x^2 - 3x - 4
Next, we can factor the quadratic equation or use the quadratic formula to find the values of x.
Using factoring:
0 = (x - 4)(x + 1)
Setting each factor to zero:
x - 4 = 0 or x + 1 = 0
Solving for x:
x = 4 or x = -1
Therefore, the two x-intercepts are x = 4 and x = -1.
To find the x-intercepts of the equation y = x^2 - 3x - 4, we need to solve the equation for x when y is equal to zero.
In other words, we need to find the values of x for which the equation equals zero.
To do this, we set the equation equal to zero and solve for x:
0 = x^2 - 3x - 4
Now, we can use various methods to solve this quadratic equation: factoring, completing the square, or using the quadratic formula.
Let's solve it by factoring:
(x + 1)(x - 4) = 0
To find the x-intercepts, we set each factor equal to zero:
x + 1 = 0 or x - 4 = 0
Solving these equations, we find:
x = -1 or x = 4
Hence, the x-intercepts of the equation y = x^2 - 3x - 4 are x = -1 and x = 4.