What is the equation of a quadratic function whose graph passes through $(2,0)$ , $(5,0)$ , and $(3,-8)$ ? Write your equation in standard form.

facas9
2021-10-24
Answered

What is the equation of a quadratic function whose graph passes through $(2,0)$ , $(5,0)$ , and $(3,-8)$ ? Write your equation in standard form.

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okomgcae

Answered 2021-10-25
Author has **93** answers

Consider the given:

The graph passes through$(2,0)$ , $(5,0)$ and $(3,-8)$ .

Let the equation is$y=a{x}^{2}+bx+c$ .

Consider the point$(2,0)$ ,

$a{\left(2\right)}^{2}+b\left(2\right)+c=0$

$4a+2b+c=0$ ... ... (1)

Consider the point$(5,0)$ ,

$a{\left(5\right)}^{2}+b\left(5\right)+c=0$

$25a+5b+c=0$ ... ... (2)

Consider the point$(3,-8)$ ,

$a{\left(3\right)}^{2}+b\left(3\right)+c=-8$

$9a+3b+c=-8$ ... ... (3)

From equation (1) and (2),

$4a+2b+c=25a+5b+c$

$4a+2b=25a+5b$

$a\u201325a=5b\u20132b$

$-7a=b$

From equation (1),

$4a+2b+c=0$

$4a+2(-7a)+c=0$

$4a\u201314a+c=0$

$-10a+c=0$

$c=10a$

Substitute the value of b and c in equation (3).

$9a+3b+c=-8$

$9a+3(-7a)+\left(10a\right)=-8$

$9a\u201321a+10a=-8$

$-21a+19a=-8$

$-2a=-8$

$a=4$

$b=-7a$

$b=-7\left(4\right)$

$b=-28$

$c=10a$

$c=10\left(4\right)$

$c=40$

Hence, The equation of quadratic function will be:

$y=4{x}^{2}\u201328x+40$

The graph passes through

Let the equation is

Consider the point

Consider the point

Consider the point

From equation (1) and (2),

From equation (1),

Substitute the value of b and c in equation (3).

Hence, The equation of quadratic function will be:

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Find the absolute maximum and absolute minimum values of f on the given interval.

$f\left(x\right)=2{x}^{3}-3{x}^{2}-36x+5,[-3,4]$

absolute minimum value

absolute maximum value

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Suppose we have a $1$-dimensional differential inequality

$\frac{dx}{dt}\le x-{x}^{3}$

We can apply the Comparison principle to claim that if $y(t)$ is the solution to $\frac{dy}{dt}=y-{y}^{3}$, then $x(t)\le y(t)$ (assuming $x(0)\le y(0)$. Can we extend similar argument to a $2$-dimensional system? For example, let us consider the following system of equations

$\frac{d{x}_{1}}{dt}={x}_{1}-{x}_{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{d{x}_{2}}{dt}\le {x}_{1}+{x}_{2}-\frac{{x}_{2}^{4}}{{x}_{1}^{2}}$

Is the solution to following

$\frac{d{y}_{1}}{dt}={y}_{1}-{y}_{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{d{y}_{2}}{dt}={y}_{1}+{y}_{2}-\frac{{y}_{2}^{4}}{{y}_{1}^{2}}$

related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that ${x}_{2}(t)\le {y}_{2}(t)$ and then subsequently use it to claim ${x}_{1}(t)\ge {y}_{1}(t)$?

$\frac{dx}{dt}\le x-{x}^{3}$

We can apply the Comparison principle to claim that if $y(t)$ is the solution to $\frac{dy}{dt}=y-{y}^{3}$, then $x(t)\le y(t)$ (assuming $x(0)\le y(0)$. Can we extend similar argument to a $2$-dimensional system? For example, let us consider the following system of equations

$\frac{d{x}_{1}}{dt}={x}_{1}-{x}_{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{d{x}_{2}}{dt}\le {x}_{1}+{x}_{2}-\frac{{x}_{2}^{4}}{{x}_{1}^{2}}$

Is the solution to following

$\frac{d{y}_{1}}{dt}={y}_{1}-{y}_{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\frac{d{y}_{2}}{dt}={y}_{1}+{y}_{2}-\frac{{y}_{2}^{4}}{{y}_{1}^{2}}$

related with the solution of the original problem. Specifically, can we apply the Comparison principle to first say that ${x}_{2}(t)\le {y}_{2}(t)$ and then subsequently use it to claim ${x}_{1}(t)\ge {y}_{1}(t)$?

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Is it true that vector spaces are defined to check if system of linear equations is solvable or not?

Explanation: Goal is to solve system of linear equations.

In matrix form:$X=b$ . As $A=[{C}_{1}{C}_{2}\dots {C}_{n}]$ , where $C}_{n$ is a column and $x=[{x}_{1}{x}_{2}\dots {x}_{n}]$ .

Therefore,${C}_{1}{x}_{1}+{c}_{2}{x}_{2}+\dots +{C}_{n}{x}_{n}=b$ . Linear combination of column vectors produce vector b.

Because of above statement (linear combination) we choose a set of vectors that have closure under addition and scalar multiplication (closure under linear combination) and call that set of vectors a vector space. Now, if vector b lies in that set of vectors (vector space) then only system of linear equations is solvable.

Explanation: Goal is to solve system of linear equations.

In matrix form:

Therefore,

Because of above statement (linear combination) we choose a set of vectors that have closure under addition and scalar multiplication (closure under linear combination) and call that set of vectors a vector space. Now, if vector b lies in that set of vectors (vector space) then only system of linear equations is solvable.

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Solutions to

${x}^{3}+{y}^{3}+{z}^{3}={x}^{2}+{y}^{2}+{z}^{2}=x+y+z=0$

I need to prove that $xyz=1$. dealing trew this problem I get that $x+y=\frac{2}{3}$ and that results that $z=-\frac{2}{3}$. after all i got $xyz=-\frac{10}{27}$

When I was before dealing with this, I over and over get that $xyz=0$.

${x}^{3}+{y}^{3}+{z}^{3}={x}^{2}+{y}^{2}+{z}^{2}=x+y+z=0$

I need to prove that $xyz=1$. dealing trew this problem I get that $x+y=\frac{2}{3}$ and that results that $z=-\frac{2}{3}$. after all i got $xyz=-\frac{10}{27}$

When I was before dealing with this, I over and over get that $xyz=0$.