Write the equation of the quadratic function with x-intercepts at (−1,0) and (5,0), and y-intercept at (0,−15). NEED HELP ASAP!!!
i got u
hold lemme do it really quick
x-intercepts at (−1,0) and (5,0)
y = a(x+1)(x-5)
y(0) = -15, so
a(1)(-5) = -15
a = 3
so y = 3(x+1)(x-5)
To write the equation of a quadratic function, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
Given that the x-intercepts are (-1,0) and (5,0), we know that the function crosses the x-axis at x = -1 and x = 5. This means that these x-values are the roots of the quadratic equation.
The factored form of the quadratic equation will be:
(x - x1)(x - x2) = 0
where x1 and x2 are the x-intercepts. Substituting the given x-intercepts, we get:
(x - (-1))(x - 5) = 0
Expanding this equation, we have:
(x + 1)(x - 5) = 0
(x^2 - 4x - 5) = 0
Now, to find the value of 'a' in the general form of the quadratic equation (ax^2 + bx + c = 0), we can use the y-intercept (0, -15). Plug in these values into the equation:
a(0)^2 + b(0) + c = -15
This simplifies to:
c = -15
So, with 'c' determined, our quadratic equation becomes:
x^2 - 4x - 5 = 0
This is the equation of the quadratic function with x-intercepts at (-1, 0) and (5, 0), and the y-intercept at (0, -15).