Question

1. consider the quadratic function ( f(x)=(x - 2)^2 ).

3c graph the parabola corresponding to ( f(x) ) on the axes below:

Explanation:

Step1: Identify the vertex form of a quadratic function

The vertex form of a quadratic function is \(y = a(x - h)^2 + k\), where \((h,k)\) is the vertex. For \(f(x)=(x - 2)^2\), \(a = 1\), \(h=2\), \(k = 0\). So the vertex is \((2,0)\).

Step2: Find the y - intercept

Set \(x = 0\). Then \(f(0)=(0 - 2)^2=4\). So the y - intercept is \((0,4)\).

Step3: Find another point

Set \(x=1\). Then \(f(1)=(1 - 2)^2 = 1\). So the point is \((1,1)\).

Step4: Use symmetry

Since the parabola \(y=(x - 2)^2\) is symmetric about the line \(x = 2\). If \((x_1,y_1)\) is a point on the parabola, then \((4 - x_1,y_1)\) is also a point on the parabola. For example, if \(x_1=1\), then \(4 - x_1 = 3\) and \(f(3)=(3 - 2)^2=1\).

Plot the vertex \((2,0)\), the y - intercept \((0,4)\), the point \((1,1)\) and its symmetric point \((3,1)\) and draw a smooth curve passing through these points.

Answer:

The parabola with vertex \((2,0)\), passing through \((0,4)\), \((1,1)\) and \((3,1)\) is the graph of \(y=(x - 2)^2\).