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4 years ago

Fit a quadratic function to these three points: (-2,8)(0,-4),and (4,68)

1 answer:

6 0

Given three points
P1(-2,8)
P2(0,-4)
P3(4,68)
We need the quadratic equation that passes through all three points.

Solution:
We first assume the final equation to be
f(x)=ax^2+bx+c .............................(0)

Observations:
1. Points are not symmetric, so cannot find vertex visually.
2. Using the point (0,-4) we substitute x=0 into f(x) to get
f(0)=0+0+c=-4, hence c=-4.
3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.
f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)
f(4)=a(4^2)+b(4)-4=68 => 16a+4b-4=68.............(2)
4. Solve system
2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)
substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2 ..........(4)
5. Put values c=-4, a=4, b=2 into equation (0) to get

f(x)=4x^2+2x-4

Check:
f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8
f(0)=0+0-4 = -4
f(4)=4(4^2)+2(4)-4=64+8-4=68
So all consistent, => solution ok.

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<h3>Explamation :</h3>

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Answer:

Verified!

Step-by-step explanation:

Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.

Let's apply it for 2x2 matix:

$A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0$

So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)$ .

So, eigenvalues are $\lambda_1=a,\:\lambda_2=c$

Let's apply it for 3x3 matrix:

$A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0$

So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have $(\lambda-a)(\lambda-c)(\lambda-f)$ .

So eigenvalues are $\lambda_1=a,\:\lambda_2=c,\:\lambda_3=f$

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