f(x) is a polynomial & f(-1) = 4. f(1) = 3. f(4) = 2. Then the graph of f(x) must cross the x-axis?

Question

3 times, is that right

Answer 1

No, the graph need not cross the x-axis. To find a counterexample, we can draw the Lagrange interpolating polynomial [we'll call it p(x)] around those three points, which is the polynomial of power two that passes through these points.

For 3 points, the polynomial is

y1*(x-x2)(x-x3)/(x1-x2)(x1-x3) +
y2*(x-x1)(x-x3)/(x2-x1)(x2-x3) +
y3*(x-x1)(x-x2)/(x3-x1)(x3-x2)

= 4(x-1)(x-4)/(-2)(-5) + 3(x+1)(x-4)/(2)(-3) +
2(x+1)(x-1)/(5)(3)

=0.4(x^2 - 5x +4) -0.5(x^2 -3x -4) + (2/15) (x^2 - 1)

so, p(x) = 1/30 x^2 -.5x +3.466666667.

By the quadratic formula, the zeros are
.5 +- sqrt(.25 - .4622222222)
--------------------------------------
2/30

Which has no real solutions, so p(x) is never zero. Since polynomials are continuous, p(x) is always positive. So f(x) need not cross the x-axis. Of course, it would be possible to create a polynomial that passes through these points and does cross the x-axis. We'd just have to add a point below the x-axis and find the lagrange interpolating polynomial that passes through those four points.

Answer 2

Previous answerer is correct. In fact, the three points you have listed are consistent with the function:

f(x) = (2x^2 - 30x + 208)/60

This function *never* crosses the x-axis. You can check this for yourself by finding the roots of 2x^2 - 30x + 208, i.e., find the values of x for with f(x)=0. You'll find that this quadratic has no real roots, and thus f(x) never crosses (or even touches) the x-axis.

There are an infinite number of polynomial functions that pass through these three points. Some may cross the x-axis, but others may not.

Answer 3

from the information given it is impossible to tell how many time the graph would touch the X axis. the information indicate that the graph is above the x axis at all 3 point. there is a chance that the graph would not even touch the axis. them again it might touch it 1, 2, 3, 100? depending on the function itself

Answer 4

not necessarily.... from the three points you have given, all of them are above the x axis... the graph could cross the x-axis, or may not cross it all... depending on the order of the polynomial

Answer 5

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