find the zeros of the function by rewriting the function in intercept form?
y=81x^2-16
y=(9x-4)(9x+4)
Now when is y equal to zero?
is it -16
To find the zeros of the function by rewriting it in intercept form, we set y to equal zero and solve for x. The intercept form of an equation is written as x = a, where "a" is the x-intercept.
Let's rewrite the equation:
0 = 81x^2 - 16
To remove the constant term (-16), we'll add 16 to both sides:
16 = 81x^2
Next, divide both sides by 81:
16/81 = x^2
To eliminate the squared term, we take the square root of both sides (remembering to consider both the positive and negative roots):
x = ±√(16/81)
Simplifying the square root:
x = ±(4/9)
Therefore, the zeros of the function, when rewritten in intercept form, are x = 4/9 and x = -4/9.
To find the zeros of the function y = 81x^2 - 16 by rewriting it in intercept form, we need to set y equal to zero and solve for x.
In the intercept form, the equation is written as (x - a)(x - b) = 0, where a and b are the x-intercepts (or zeros) of the function.
So, setting y = 0 in the given equation:
0 = 81x^2 - 16
Next, we can factor the equation by using the difference of squares formula, which states that a^2 - b^2 can be factored as (a - b)(a + b).
In this case, the equation can be rewritten as:
0 = (9x)^2 - 4^2
Using the difference of squares, we get:
0 = (9x - 4)(9x + 4)
Now that we have factored the equation, we can set each factor equal to zero and solve for x:
9x - 4 = 0
9x = 4
x = 4/9
and
9x + 4 = 0
9x = -4
x = -4/9
Therefore, the zeros of the function y = 81x^2 - 16 written in intercept form are x = 4/9 and x = -4/9.