Find the value of ##a, b## and ##k## in the problem involving graphs

[chwala]

Homework Statement

See attached

Relevant Equations

Graphs

In my approach i have the roots of the equation being $x=a$ and $x=b$.

There are two assumptions,
In the first assumption,

$a=\dfrac{1}{2}b$

$2a=b$

then,

$4=k(-a)^2(-2a)$

$4=-2ka^3$

$⇒ -2=ka^3$

Now since $2a=b$ then $a=1, b=2⇒k=-2$.

our equation becomes,

$f(x)=-2(x-1)^2(x-2)$

checking using,

$f(1.5)=-2(1.5-1)^2(1.5-2)=0.25>0$

Now to the second assumption,
Let

$b=\dfrac{1}{2}a$

$2b=a$

then,

$4=k(-2b)^2(-b)$

$4=-4kb^3$

$⇒ -1=kb^3$

Now since $2b=a$ then $b=1, a=2 ⇒k=-1$.

Our equation becomes,

$f(x)=-1(x-2)^2(x-1)$

checking using,

$f(1.5)=-2(1.5-1)^2(1.5-2)=-0.125 <0$ which is a contradiction as $y$ is positive between the roots $x=1$ and $x=2$.
Therefore the second assumption does not apply. We shall therefore have the unknown values given by:

$a=1, b=2, k=-2$.

There may be a better approach.

 

[topsquark]

Homework Statement: See attached
Relevant Equations: Graphs

View attachment 335943

In my approach i have the roots of the equation being $x=a$ and $x=b$.

There are two possibilities,
In the first possibility,

$a=\dfrac{1}{2}b$

$2a=b$

then,

$4=k(-a)^2(-2a)$

$4=-2ka^3$

$⇒ -2=ka^3$

Now since $2a=b$ then $a=1, b=2⇒k=-2$.

our equation becomes,

$f(x)=-2(x-1)^2(x-2)$

checking using,

$f(1.5)=-2(1.5-1)^2(1.5-2)=0.25>0$

Now to the second possibility,
Let

$b=\dfrac{1}{2}a$

$2b=a$

then,

$4=k(-2b)^2(-b)$

$4=-4kb^3$

$⇒ -1=kb^3$

Now since $2b=a$ then $b=1, a=2 ⇒k=-1$.

Our equation becomes,

$f(x)=-1(x-2)^2(x-1)$

checking using,

$f(1.5)=-2(1.5-1)^2(1.5-2)=-0.125 <0$ which is a contradiction as $y$ is positive between the roots $x=1$ and $x=2$.

Therefore,

$a=1, b=2, k=-2$.

There may be a better approach.

Did you graph your answers to see which is right?

Hint: Double roots "touch" the x-axis, whereas single roots cross it.

Which is your double root?

-Dan

 

[chwala]

Did you graph your answers to see which is right?

Hint: Double roots "touch" the x-axis, whereas single roots cross it.

Which is your double root?

-Dan

I think i have shown that the second possibility does not apply thus we can only have the first possibility as the correct approach.
Kindly follow my working.

 

[PeroK]

I think i have shown that the second possibility does not apply thus we can only have the first possibility as the correct approach.
Kindly follow my working.

You asked for a better method. What Dan's saying is that $\dfrac{dy}{dx}(x=a) = 0$, so the double root at $x = a$ is a turning point. That gives you $a = 1$ directly.

 

[Mark44]

@chwala, your solution is much more long-winded than is necessary. One can see by inspection of the graph that a = 1 and b = 2. The equation is therefore $y = k(x - 1)^2(x - 2)$.
Since the y-intercept is 4, k can be determined by solving the equation 4 = k(-1)(-2).

 

[chwala]

@chwala, your solution is much more long-winded than is necessary. One can see by inspection of the graph that a = 1 and b = 2. The equation is therefore $y = k(x - 1)^2(x - 2)$.
Since the y-intercept is 4, k can be determined by solving the equation 4 = k(-1)(-2).

I do not think it was that straightforward. How to answer the question on script (by inspection )if it was an exam question? ...using complete square knowledge to justify $a$ as a turning point of graph I suppose.
Cheers man.

 

[Mark44]

I do not think it was that straightforward. How to answer the question on script (by inspection )if it was an exam question?

It was straightforward to me. You were given the equation $y = k(x - a)^2(x - b)$ and shown that the graph has x-intercepts at x = 1 and x = 2. I'm assuming that these are the exact values, and I'm sure that was the intent of the writer of this problem.

The facts that 1) the graph had a parabolic shape that opened upward near x = 1 and 2) crossed the x-axis at x = 2, told me that the quadratic factor had to be $(x - 1)^2$ and that the linear factor had to be $(x - 2)$. Being able to recognize the behavior of low-degree polynomials is a skill that is often taught as part of precalculus courses.

.using complete square knowledge to justify a as a turning point of graph I suppose.

Completing the square, if that's what you meant above, isn't helpful. The equation is already given as a product of factors.