A quartic polynomial has $2$ distinct real roots at $x=1$ and $x=-3/5$. If the function has a $y$-intercept at $-1$ and has $f(2)=2$ and $f(3)=3$. How to determine the remaining roots when two distinct real roots and y-intercept are given.

Khalfanebw
2022-09-19
Answered

A quartic polynomial has $2$ distinct real roots at $x=1$ and $x=-3/5$. If the function has a $y$-intercept at $-1$ and has $f(2)=2$ and $f(3)=3$. How to determine the remaining roots when two distinct real roots and y-intercept are given.

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Mackenzie Lutz

Answered 2022-09-20
Author has **2** answers

You can sum up everything you know in a system like this:

$\{\begin{array}{l}f(1)=0\\ f(-\frac{3}{5})=0\\ f(0)=-1\\ f(2)=2\\ f(3)=3\end{array}$

Then it's very simple to find all the coefficients of $f(x)=a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e$, because you have a $5$-variable system and $5$ equations, so the solution is unique! For example, from the $3$rd equation we get $f(0)=0+0+0+0+e=-1\Rightarrow e=-1$, and so on. After some calculus we obtain:

$\{\begin{array}{l}a=\frac{5}{156}\\ b=-\frac{41}{78}\\ c=\frac{289}{156}\\ d=-\frac{14}{39}\\ e=-1\end{array}$

Finally we manage to find the polynomail, and the associated function:

$f(x)=\frac{5}{156}{x}^{4}-\frac{41}{78}{x}^{3}+\frac{289}{156}{x}^{2}-\frac{14}{39}x-1$

So all the real roots are ${x}_{1}=-\frac{3}{5},{x}_{2}=1,{x}_{3}=8-2\sqrt{3},{x}_{4}=8+2\sqrt{3}$.

$\{\begin{array}{l}f(1)=0\\ f(-\frac{3}{5})=0\\ f(0)=-1\\ f(2)=2\\ f(3)=3\end{array}$

Then it's very simple to find all the coefficients of $f(x)=a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e$, because you have a $5$-variable system and $5$ equations, so the solution is unique! For example, from the $3$rd equation we get $f(0)=0+0+0+0+e=-1\Rightarrow e=-1$, and so on. After some calculus we obtain:

$\{\begin{array}{l}a=\frac{5}{156}\\ b=-\frac{41}{78}\\ c=\frac{289}{156}\\ d=-\frac{14}{39}\\ e=-1\end{array}$

Finally we manage to find the polynomail, and the associated function:

$f(x)=\frac{5}{156}{x}^{4}-\frac{41}{78}{x}^{3}+\frac{289}{156}{x}^{2}-\frac{14}{39}x-1$

So all the real roots are ${x}_{1}=-\frac{3}{5},{x}_{2}=1,{x}_{3}=8-2\sqrt{3},{x}_{4}=8+2\sqrt{3}$.

asked 2022-06-15

Find the $y$-intercept of a given parametric curve.

$x=t({t}^{2}-3)$

$y=3({t}^{2}-3)$

$x=t({t}^{2}-3)$

$y=3({t}^{2}-3)$

asked 2022-06-26

Two parallel lines, the equation of the first line is

$y=-x-3.$

The distance between the parallel lines is $0.5$. How do I find the $y$-intercept of the second line?

$y=-x-3.$

The distance between the parallel lines is $0.5$. How do I find the $y$-intercept of the second line?

asked 2022-07-03

I know how to find the parabola quadratic equation given the roots. However, in this problem I'm given the y-intercept of (0,3) {So, now I know C value} and the axis of symmetry of x=-3/8. From this I know that 3a=4b. But, it seems I need one more point, since I have 3 unknowns, I'd need 3 INDEPENDENT equations.

So, I said, ah, I know another point, it's on the other side of the aos. So, I used (-3/4, 3). However, when I plug this in, it appears to not be an independent equation. So, I'm left with just c=3 and 3a=4b. How, do I work this from here?

So, I said, ah, I know another point, it's on the other side of the aos. So, I used (-3/4, 3). However, when I plug this in, it appears to not be an independent equation. So, I'm left with just c=3 and 3a=4b. How, do I work this from here?

asked 2022-06-27

Function: $f(x)=\frac{\mathrm{sin}(x)}{x}$. Why is the $y$-intercept $=1$ despite $\mathrm{sin}(0)$ being divided by zero?

asked 2022-05-09

Find the $y$-intercept of the curve that passes through the point $(2,1)$ with the slope at $(x,y)$ of $\frac{-9}{{y}^{2}}$

$\frac{dy}{dx}=\frac{-9}{{y}^{2}}$

$\int {y}^{2}dy=\int -9dx$

$\frac{{y}^{3}}{3}=-9x+{C}_{1}$

${y}^{3}=-27x+C$ $(C=3{C}_{1})$

$y=(-27x+C{)}^{1/3}$

$1=(-27(2)+C{)}^{1/3}$

$C=54$

$0=(-27x+54{)}^{1/3}$

$y-intercept=(-2,0)$

Where is mistake?

$\frac{dy}{dx}=\frac{-9}{{y}^{2}}$

$\int {y}^{2}dy=\int -9dx$

$\frac{{y}^{3}}{3}=-9x+{C}_{1}$

${y}^{3}=-27x+C$ $(C=3{C}_{1})$

$y=(-27x+C{)}^{1/3}$

$1=(-27(2)+C{)}^{1/3}$

$C=54$

$0=(-27x+54{)}^{1/3}$

$y-intercept=(-2,0)$

Where is mistake?

asked 2022-06-14

Consider the surface given by the function $f(x,y)=\frac{4}{\sqrt{{x}^{2}+{y}^{2}-9}}$. What is the range of $f$? Determine any $y$-intercept(s) of $f$.

For range, I realize that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number, so long as ${x}^{2}+{y}^{2}-9$ is positive or equal to zero. However, when looking at $f$, we know that the denominator cannot equal zero, which means that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number greater than $0$. Would this mean that the range of $f$ is $(0,\mathrm{\infty})$?

For the $y$-intercept, I know that both $x$ and $z$ need to equal zero. Would this mean that there are no $y$-intercepts because $0$ is not in the range of $f$?

For range, I realize that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number, so long as ${x}^{2}+{y}^{2}-9$ is positive or equal to zero. However, when looking at $f$, we know that the denominator cannot equal zero, which means that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number greater than $0$. Would this mean that the range of $f$ is $(0,\mathrm{\infty})$?

For the $y$-intercept, I know that both $x$ and $z$ need to equal zero. Would this mean that there are no $y$-intercepts because $0$ is not in the range of $f$?

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What is the slope of a line that has an X-Intercept of 8 and a Y-Intercept of 11?