Question

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short answer

1. graph the circle. $(x - 4)^2+(y + 3)^2 = 9$
2. graph the absolute value function $g(x)=\frac{2}{3}|x - 1|+2$
3. graph the quadratic function $g(x)=2(x + 3)^2-6$

Explanation:

Step1: Identify the center and radius of the circle

The standard form of a circle is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.
For the equation \((x - 4)^2+(y + 3)^2 = 9\), we have \(h = 4\), \(k=-3\), and \(r = 3\) (since \(r^2=9\)).

Step2: Plot the center

Plot the point \((4,-3)\) on the coordinate plane.

Step3: Plot points around the center at a distance of the radius

From the center \((4,-3)\), move 3 units up (to \((4,0)\)), 3 units down (to \((4,-6)\)), 3 units left (to \((1,-3)\)), and 3 units right (to \((7,-3)\)). Then sketch the circle passing through these points.

Step4: Analyze the absolute - value function \(g(x)=\frac{2}{3}|x - 1|+2\)

The parent function is \(y = |x|\). The transformation \(y=a|x - h|+k\) has a vertex at \((h,k)\) and a vertical stretch or compression factor \(|a|\).
For \(g(x)=\frac{2}{3}|x - 1|+2\), the vertex is \((1,2)\) and the vertical compression factor is \(\frac{2}{3}\).

Step5: Plot points for the absolute - value function

When \(x = 1\), \(y = 2\) (vertex).
When \(x=4\), \(y=\frac{2}{3}|4 - 1|+2=\frac{2}{3}\times3+2=4\).
When \(x=-2\), \(y=\frac{2}{3}|-2 - 1|+2=\frac{2}{3}\times3+2 = 4\).
Plot the vertex \((1,2)\) and the points \((4,4)\) and \((-2,4)\) and then sketch the V - shaped graph of the absolute - value function.

Step6: Analyze the quadratic function \(g(x)=2(x + 3)^2-6\)

The parent function is \(y=x^2\). The transformation \(y=a(x - h)^2+k\) has a vertex at \((h,k)\) and a vertical stretch or compression factor \(|a|\).
For \(g(x)=2(x + 3)^2-6\), the vertex is \((-3,-6)\) and the vertical stretch factor is \(2\).

Step7: Plot points for the quadratic function

When \(x=-3\), \(y=-6\) (vertex).
When \(x=-2\), \(y=2(-2 + 3)^2-6=2\times1-6=-4\).
When \(x=-4\), \(y=2(-4 + 3)^2-6=2\times1-6=-4\).
When \(x=-1\), \(y=2(-1 + 3)^2-6=2\times4-6 = 2\).
When \(x=-5\), \(y=2(-5 + 3)^2-6=2\times4-6=2\).
Plot the vertex \((-3,-6)\) and the other points \((-2,-4),(-4,-4),(-1,2),(-5,2)\) and then sketch the parabola opening upwards.

Answer:

For the circle \((x - 4)^2+(y + 3)^2 = 9\): center \((4,-3)\), radius \(3\).
For the absolute - value function \(g(x)=\frac{2}{3}|x - 1|+2\): vertex \((1,2)\), vertical compression by \(\frac{2}{3}\).
For the quadratic function \(g(x)=2(x + 3)^2-6\): vertex \((-3,-6)\), vertical stretch by \(2\). Graphs are sketched according to the above - mentioned key points.