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QUESTION IMAGE

find the coordinates of the vertices of the image of math for each tran…

Question

find the coordinates of the vertices of the image of math for each transformation.

  1. translation $(x,y)\to(x - 1,y + 2)$

$m$ ( type your answer... type your answer... ) and $a$ ( type your answer... type your answer... )

  1. reflection across the $x$-axis

$m$ ( type your answer... type your answer... ) and $t$ ( type your answer... type your answer... )

  1. counter-clockwise rotation 90 degrees about point $(0,0)$

$m$ ( type your answer... type your answer... ) and $h$ ( type your answer... type your answer... )

  1. dilation with a scale factor of 4

$m$ ( type your answer... type your answer... ) $h$ ( type your answer... type your answer... )

  1. apply the sequence with translation $(x,y)\to(x - 4,y)$ and reflection across the line $y=-2$

$m$ ( type your answer... type your answer... ) $m$ ( type your answer... type your answer... )

Explanation:

Step1: Find original coordinates

From the graph, \(M(-2,2)\), \(A(3,0)\), \(T(0,-4)\), \(H(-3,-2)\)

Step2: Solve for translation \((x,y)\to(x - 1,y+2)\)

For \(M\): \(x=-2-1=-3\), \(y = 2+2=4\), so \(M'(-3,4)\)
For \(A\): \(x=3-1=2\), \(y=0 + 2=2\), so \(A'(2,2)\)

Step3: Solve for reflection across the \(x\) - axis \((x,y)\to(x,-y)\)

For \(M\): \(x=-2\), \(y=-2\), so \(M'(-2,-2)\)
For \(T\): \(x = 0\), \(y = 4\), so \(T'(0,4)\)

Step4: Solve for counter - clockwise rotation \(90^{\circ}\) about \((0,0)\) \((x,y)\to(-y,x)\)

For \(M\): \(x=-2\), \(y=-2\), so \(M'(-2,-2)\)
For \(H\): \(x = 2\), \(y=-3\), so \(H'(2,-3)\)

Step5: Solve for dilation with scale factor \(4\) \((x,y)\to(4x,4y)\)

For \(M\): \(x=-2\times4=-8\), \(y=2\times4 = 8\), so \(M'(-8,8)\)
For \(H\): \(x=-3\times4=-12\), \(y=-2\times4=-8\), so \(H'(-12,-8)\)

Step6: Solve for the sequence: translation \((x,y)\to(x - 4,y)\) then reflection across \(y=-2\)

First, translation for \(M(-2,2)\): \(x=-2-4=-6\), \(y = 2\), so \(M_1(-6,2)\)
Then reflection across \(y=-2\): The distance from \(y = 2\) to \(y=-2\) is \(4\). So the new \(y\) - coordinate is \(-2-4=-6\), \(x=-6\), so \(M'(-6,-6)\)

Answer:

  1. \(M'(-3,4)\), \(A'(2,2)\)
  2. \(M'(-2,-2)\), \(T'(0,4)\)
  3. \(M'(2,-2)\), \(H'(2,-3)\)
  4. \(M'(-8,8)\), \(H'(-12,-8)\)
  5. \(M'(-6,-6)\)