Question

27. het f(x) be a polynomial function, with f(b) = 3, f(c) = 0, f(d) =1, and fle) -3, and becedce. which is true ? (3 pts) a) x-c is a factor of f(x) b) xtc is a factor of f(x) c) c is likely a touch point d) c is likely a cross point e) there exists n between c and a such that f(n) = 0 f) there exists p between aande such that f(p) = 0

Answer 1

Solution:-

We have given that

f(x) is a polynomial.

f(b)= 3,

f(c)= 0,

f(d)= 1,

f(e)= -3 and

b < c < d < e

Now, let us see which option is correct.

(a)

Since x = c results in f(c)=0.

So, x - c is a factor of f(x).

Hence, option - (a) is correct.

(b)

Since x +c is a factor of polynomial if x = -c results in f(-c)=0.

But nothing is given for x = -c.

So, option (b) is incorrect.

(c)

Since b < c < d

And f(b) = 3 and f(d) = 1 both are positive which f(b) is negative.

We know that if both sides of a zero of a polynomial are of same sign(Here positive). Then the point of x zero must be a touch point on tye x-axis.

So, here, x = c is zero of f(x) and its both side f(x) is positive.

So, x = c can be a touch point. (Here we are not sure because there may be other points on the graph between b and d that affect our conclusion but still there are chances. ).

So, option c that c is liklely a touch point is correct.

Hence, option (c) is correct

(d)

Here, also we can not conclude that c is a cross point because we have given nothing that can be in favour.

So, option (d) is also incorrect.

(e)

Since f(c) =0 and f(d) = 1 & c < d.

So, it is wrong to conclude that a point n between c and d results in f(n)=0.

So, option (e) is also incorrect.

(f)

Since f(d)= 1 and f(e) = -3 and d < e.

It means from x = d to x = e, the graph of the polynomial must cross the x- axis.

So, there must be a point x = p , for which f(p) =0.

Hence, option (f) is also a correct.