I don't know how to put this in standard form:
f(x)=x^2-2x-15
I tried (x^2-2x+-x^2)-25-x^2, but x^2 doesn't factor into anything so I can't fully put it into standard form. can you show me how, plus how to find the x intercepts and y intercepts?
f(x)=(x-1)^2 -16
and of course, that is a difference of two squares...
To put a quadratic function in standard form, you'll need to rearrange it so that it is written as a trinomial in descending order of the exponents.
In this case, the given quadratic function is f(x) = x^2 - 2x - 15.
To put it in standard form, you'll start by rearranging the terms in descending order of exponents:
f(x) = x^2 - 2x - 15
Standard form: f(x) = ax^2 + bx + c, where a, b, and c are constants.
Comparing it to the standard form, we can identify a = 1, b = -2, and c = -15.
So, the standard form of the given quadratic function f(x) = x^2 - 2x - 15 is f(x) = 1x^2 - 2x - 15.
To find the x-intercepts of a quadratic function, you need to set f(x) equal to zero and solve for x:
1x^2 - 2x - 15 = 0
To factor or solve the equation, you can use various methods like factoring, quadratic formula, or completing the square. In this case, factoring is the most straightforward approach.
(x - 5)(x + 3) = 0
Setting each factor equal to zero:
x - 5 = 0 => x = 5
x + 3 = 0 => x = -3
Thus, the x-intercepts are x = 5 and x = -3.
To find the y-intercept, you need to evaluate the function when x is zero (f(0)):
f(0) = 1(0)^2 - 2(0) - 15
f(0) = -15
Therefore, the y-intercept is y = -15.