Question

The figure shows the curves represented by polynomials f(x)f(x), g(x)g(x), and h(x)h(x) of degrees 4, 4, and 2 respectively, on XYXY plane. Let f(x)-g(x)= ax(x-2)(x-5)(x-9)f(x)−g(x)=ax(x−2)(x−5)(x−9), a \neq 0a  =0. If bb is a negative constant, then choose the most possible expression for h(x)h(x) and other correct statements among the given options. (Note that figure is not according to scale .)
>h(x)=b(x ^2 +8x−7)

>f(x)=g(x) at x=0,-2,-5,-9
>h(x)= b(x^2-6x-7)


>h(x)= b(x^2-2x-3)

>h(x)= b(x^2-8x+7)

>h(x)= b(x^2-6x+7)

>f(x)=g(x) at x=0,2,5,9​

Answer 1

Possible expression of h(x) is b($x^{2} -6x-7$ ) and f(x) = g(x) for x = 0,2,5,9 Hence option 3 and option 7 is correct.

Given : f(x) is a polynomial of degree 4

            g(x) is a polynomial of degree 4

           h(x) is a polynomial of degree 2

          f(x) - g (x) is given by ax(x-2)(x-5)(x-9) where a is not equal to 0

To Find : Possible expression of h(x) and the other statements given in the options

1) h(x)=b(x ^2 +8x−7)

2)f(x)=g(x) at x=0,-2,-5,-9

3)h(x)= b(x^2-6x-7)

4)h(x)= b(x^2-2x-3)

5)h(x)= b(x^2-8x+7)

6) h(x)= b(x^2-6x+7)

7) f(x)=g(x) at x=0,2,5,9​

Solution : The correct option of the given problem is 3 and 7

h(x) = b($x^{2} -6x-7$) and f(x) = g(x) at x = 0 , 2, 5 , 9

It is given that  f(x) - g(x) is ax(x-2)(x-5)(x-9) where a is not equal to 0

So putting f(x) - g(x) = 0 we get

ax(x-2)(x-5)(x-9) = 0

this will gives  x = 0 , x-5 =0 , x-2 = 0 and x-9 =0

therefore the value of x where f(x) = g(x) is x is 0 , 2 , 5 , 9

So option 7 is correct .

Now to find the correct expression of h(x) we see that h(x) is an upward parabola , the y coordinate is greater than zero.

Let h(x) =b( $ax^{2} +bx+c$) where b is negative constant

now h(0) = bc and b is negative , y coordinate is greater than zero so c should be less than 0

i.e. c < 0

so option 5 and 6 are not possible

since c > 0 in them

Now

By the given graph it is clear that one  x coordinate of h(x) lies in between the points 5 and 9 and other coordinate is less than 0.

If h(x) is b(x ^2 +8x−7)

then h(x) = 0

b(x ^2 +8x−7) = 0

(x ^2 +8x−7) = 0

by solving the above quadratic equation we have

x = 0.79 and x = -8.79

which is not possible

So option 1 is discarded.

Now

if h(x) = b(x^2-6x-7)

h(x) = 0

b(x^2-6x-7) = 0

(x^2-6x-7) = 0

(x+1) (x-7) = 0

= x = -1 and x = 7

so the above values of x is possible

So option 3 is correct

Now if h(x) = b(x^2-2x-3)

h(x) = 0

b(x^2-2x-3) = 0

(x^2-2x-3) = 0

(x+1) (x-3) = 0

x = -1 and x = 3

Which is not possible

So option 4 is incorrect.

So possible expression of h(x) is b($x^{2} -6x-7$ ) and f(x) = g(x) for x = 0,2,5,9 Hence option 3 and option 7 is correct.

Answer 2

For x = 0,2,5,9, $\mathbf{b}\left(x^{2}-6 x-7\right)$ and f(x) = g(x) are potential expressions for h(x). As a result, options 3 and 7 are both correct.

Given: The polynomial f(x) has a degree of 4.

A polynomial of degree 4 is g(x).

A polynomial of degree two is h(x).

Given by ax(x-2)(x-5)(x-9) where an is not equal to 0, f(x) - g(x) is obtained.

Possible formulations of the other statements provided in the alternatives, as well as h(x),

1) h(x)=b(x ^2 +8x−7).

2)f(x)=g(x), where x=0,-2,-5, and 9.

3)h(x)= b(x^2-6x-7).

4)h(x)= b(x^2-2x-3).

5)h(x)= b(x^2-8x+7).

6) h(x)= b(x^2-6x+7).

7) At x=0,2,5,9, f(x)=g(x).

​The answer to the presented puzzle is 3, and 7.

h(x) =$\mathbf{b}\left(x^{2}-6 x-7\right)$ and f(x) = g(x) at x = 0 , 2, 5 , 9

Given that an is not equal to zero, f(x) - g(x) = ax(x-2)(x-5)(x-9)

Consequently, f(x) - g(x) = 0 gives us

ax(x-2)(x-5)(x-9) = 0

This results in x = 0 and x-5 = 0. , x-2 = 0 and x-9 =0

Consequently, x has the value of 0, 2, 5, or 9 when f(x) = g(x).

Therefore, choice 7 is accurate.

Now that we know that h(x) is an upward parabola and that the y coordinate is greater than zero, we can get the proper formula for h(x).

If h(x) $=b\left(a x^{2}+b x+c\right)$ When the constant b is negative

Y coordinate is higher than zero, h(0) = bc, and b is now negative, thus c should be less than 0.

i.e. c < 0

Therefore, options 5 and 6 are not viable.

because in them, c > 0,

Now,

It is evident from the following graph that one h(x) coordinate lies between the coordinates 5 and 9, whereas the other coordinate is smaller than 0.

If b(x 2 + 8x 7) is h(x),

therefore h(x) = 0

b(x ^2 +8x−7) = 0

(x ^2 +8x−7) = 0

In order to solve the quadratic equation above, we have

x = 0.79 and x = -8.79

which cannot be done

Thus, choice 1 is eliminated.

Now,

If b(x2-6x-7) = h(x), then.

h(x) = 0.

b(x^2-6x-7) = 0.

(x^2-6x-7) = 0.

(x+1) (x-7) = 0.

x = -1 and x = 7.

the x values mentioned above are thus plausible.

Option 3 is therefore the right choice.

If h(x) = b(x2-2x-3).

Now, h(x) = 0.

b(x^2-2x-3) = 0.

(x^2-2x-3) = 0.

(x+1) (x-3) = 0.

x = -1 and x = 3.

which cannot be done.

Option 4 is therefore untrue.

Hence, for x = 0,2,5,9, a possible expression of h(x) is $\mathbf{b}\left(x^{2}-6 x-7\right)$, and f(x) = g(x). As a result, options 3 and 7 are both correct.

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