HELP!!!!!!!!!!!!!
write equation of a polynomial function with the given characteristics.
Leading coefficient is 1 or -1
crosses the x axis at -2,3, and 4
touches the x axis at 0
lies above the x axis between -2 and 0
Can you write an equation of a polynomial function with given characteristics:
one x-intercept
end behaviour of Q2->Q4
y-intercept of (0,2)
To find the equation of a polynomial function with the given characteristics, we will work step-by-step:
1. Since the leading coefficient is 1 or -1, let's assume it is 1.
2. Since the polynomial crosses the x-axis at -2, 3, and 4, we know that it must have factors of (x + 2), (x - 3), and (x - 4).
3. Since the polynomial touches the x-axis at 0, we know that it has a double root of 0. Therefore, we need to include (x - 0)^2, which simplifies to x^2, as one of the factors.
4. Finally, since the polynomial lies above the x-axis between -2 and 0, we know that it should have a factor of (x + 1) which ensures the function stays positive in that interval.
Combining all the factors, we have:
f(x) = (x^2)(x + 2)(x - 3)(x - 4)(x + 1)
This is the equation of the polynomial function with the given characteristics.
To find the equation of a polynomial function with the given characteristics, we can start by using the factors of the equation.
We know that the leading coefficient is either 1 or -1, so we can say the equation has the form:
f(x) = (x - p)(x - q)(x - r)...
where p, q, r are the x-intercepts or roots of the equation.
From the given characteristics, we know that the polynomial crosses the x-axis at -2, 3, and 4. Therefore, we can write the factors as:
f(x) = (x + 2)(x - 3)(x - 4)...
Next, we are told that the polynomial also touches the x-axis at 0, which means 0 is a double root or a repeated root. So, we need to include a factor of (x - 0)^2 = x^2:
f(x) = x^2(x + 2)(x - 3)(x - 4)...
Now, we were also given that the polynomial lies above the x-axis between -2 and 0. This means there must be an even number of roots within this interval. Since we already have two roots in that range (-2 and -3), we can include an even power of (x - k), where k is any number between -2 and 0.
Let's choose k = -1 for simplicity:
f(x) = x^2(x + 2)(x - 3)(x - 4)(x + 1)²
And that's the equation of the polynomial function with the given characteristics.
from "touches the x axis at 0", we can see a double root of x=0 at the origin, meaning the curve must touch the x-axis at the origin without crossing.
But it does cross at x = -2,3 and 4
so we would have
y = ±1(x^2)(x+2)(x-3)(x-4)
[think of the x^2 as (x-0)(x-0)]
make a sketch with intercepts as I have indicated, having the curve rise to infinity in the 1st, and to negative infinity in the 3rd quadrants,
for the curve to be above the x-axis between -2 and 0
so the coefficient has to be +1
so y = (x^2)(x+2)(x-3)(x-4)