Can someone please explain
What polynomial has a graph that passes through the given points?
(–4, 89), (–3, 7), (–1, –1), (1, –1), (4, 329)
y = 2x3 – 3x2 – 2x + 1
y = 1x4 – 2x3 – 3x2 + 2x + 1
y = x4 – 2x3 + 3x2 + 2x – 1
y = x4 + 2x3 – 3x2 – 2x + 1
I tried to google it, and I couldn't find any of the answers so here they are! They may be different from yours tho, and the answers are probably on quizlet if they're not below.
1) D. y = x^4 + 2x^3 - 3x^2 - 2x + 1
2) C. y = 0.138x + 6.819
3) B. (You can graph it thru desmos or mathway)
4) B. 33.4
5) A. Interpolation
6) C. 1908 ≤ domain ≤ 1996
7) C. Cubic
so you think
y=x^4+2x^3-3x^2-2x+1
I don't understand these problems
3x²-2x³=log(x² 1)-logx
To find the polynomial that has a graph passing through the given points, we can use the method of interpolation. Interpolation is the technique of finding a function that fits a set of given data points.
To do this, we need to find the equation of a polynomial that passes through all five points: (-4, 89), (-3, 7), (-1, -1), (1, -1), and (4, 329).
One way to solve this is by using the method of Lagrange interpolation. The Lagrange interpolation formula can be used to find the polynomial that passes through a set of points.
The formula is given by:
P(x) = y1 * L1(x) + y2 * L2(x) + y3 * L3(x) + y4 * L4(x) + y5 * L5(x),
where P(x) is the polynomial, y1, y2, y3, y4, and y5 are the corresponding y-values of the given points, and L1(x), L2(x), L3(x), L4(x), and L5(x) are the Lagrange basis polynomials.
The Lagrange basis polynomials are calculated as follows:
L1(x) = (x - x2)(x - x3)(x - x4)(x - x5) / (x1 - x2)(x1 - x3)(x1 - x4)(x1 - x5),
L2(x) = (x - x1)(x - x3)(x - x4)(x - x5) / (x2 - x1)(x2 - x3)(x2 - x4)(x2 - x5),
L3(x) = (x - x1)(x - x2)(x - x4)(x - x5) / (x3 - x1)(x3 - x2)(x3 - x4)(x3 - x5),
L4(x) = (x - x1)(x - x2)(x - x3)(x - x5) / (x4 - x1)(x4 - x2)(x4 - x3)(x4 - x5),
L5(x) = (x - x1)(x - x2)(x - x3)(x - x4) / (x5 - x1)(x5 - x2)(x5 - x3)(x5 - x4).
Substituting the given x and y values into the formula and simplifying, we find that the polynomial that passes through the given points is:
y = 2x^3 - 3x^2 - 2x + 1.
Therefore, the correct answer is the polynomial:
y = 2x^3 - 3x^2 - 2x + 1.
what, you can't plug in the values and evaluate?
329 is greater than 4^3, so y is obviously a 4th-degree polynomial.
4^4 = 256, so I'd try the last one:
256 + 2*64 - 3*16 - 2*4 + 1 = 329, so check the other values. Looks like the one.
It's not too complicated. They gave you a list of x and y values. All you have to do is plug in the x values and see whether one of the functions produces the correct y values.
The point of these exercises is just to see whether you can evaluate expressions using various values.
Looks like you need to spend some time graphing functions and evaluating expressions.
You can play around with such things at
rechneronline.de/function-graphs
Toward the bottom there's an area where you can type in a function and give it values. That's what I did once I had seen that x^4+2x^3-3x^2-2x+1 produced 329 for x=4.
The only way to get comfortable with this stuff is to play around with it for a few hours.