-The quadratic coefficients is

-The linear coefficients is

-The constant term is

Cabiolab
2020-12-22
Answered

A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function.

-The quadratic coefficients is

-The linear coefficients is

-The constant term is

-The quadratic coefficients is

-The linear coefficients is

-The constant term is

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Bella

Answered 2020-12-23
Author has **81** answers

Standard vertex form of quadratic function is

Where point (h,k) is vertex point.

Also function pass through point (−2,−3).

=-1

Substitute 'a' value in equation (1)

Hence quadratic function is

Where

Quadratic coefficient =−1

Linear coefficient=2

Constant term

asked 2022-06-25

So my course shows me three differential equations:

$\dot{x}+{x}^{2}=t$

$\dot{x}=({t}^{2}+1)(x-1)$

$\dot{x}+x={t}^{2}$

The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.

Unfortunately, they don't show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?

For the second one, I thought I bring it into standard form somehow:

$\dot{x}=({t}^{2}+1)(x-1)=x{t}^{2}+x-{t}^{2}-1=x({t}^{2}+1)-{t}^{2}-1$

If we say we let $p(t)={t}^{2}+1$ and $q(t)=1+{t}^{2}$, then we could say:

$\dot{x}=xp(t)-q(t)=...$

And so on, to simplify until we reach standard form of a linear ODE (or not).

Is that the way to go? Or is there some other way to check if a ODE is linear?

$\dot{x}+{x}^{2}=t$

$\dot{x}=({t}^{2}+1)(x-1)$

$\dot{x}+x={t}^{2}$

The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.

Unfortunately, they don't show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?

For the second one, I thought I bring it into standard form somehow:

$\dot{x}=({t}^{2}+1)(x-1)=x{t}^{2}+x-{t}^{2}-1=x({t}^{2}+1)-{t}^{2}-1$

If we say we let $p(t)={t}^{2}+1$ and $q(t)=1+{t}^{2}$, then we could say:

$\dot{x}=xp(t)-q(t)=...$

And so on, to simplify until we reach standard form of a linear ODE (or not).

Is that the way to go? Or is there some other way to check if a ODE is linear?

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