Graph the equation. y=2x^2-8x+3y=2x 2 −8x+3

Mathematics

1 answer:

Gwar [14]3 years ago

8 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

The equation can be rewritten to vertex form, so you can tell the vertex is at (2, -5). The leading coefficient of 2 means the points will be twice as far apart vertically than they would be if the leading coefficient were 1. A couple of points are shown in the attached.

y = 2(x^2 -4x) +3

y = 2(x^2 -4x +4) +3 -8 . . . . . . complete the square

y = 2(x -2)^2 -5 . . . . . . . . . . . . finish putting in vertex form

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If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$