Question

sketch the graph of the quadratic function $h(x) = 3x^2 - 12x + 9$ on the coordinate plane, ensuring to clearly label the vertex, axis of symmetry, x-intercepts, y-intercept, and at least two other distinct points that demonstrate the parabolas shape, and indicate whether it opens upward or downward.

Explanation:

Step1: Determine direction of opening

The coefficient of \(x^2\) is \(3\) (positive), so the parabola opens upward.

Step2: Find axis of symmetry

For \(h(x)=ax^2 + bx + c\), axis of symmetry is \(x = -\frac{b}{2a}\). Here, \(a = 3\), \(b=-12\), so \(x=-\frac{-12}{2\times3}=2\).

Step3: Find vertex

Substitute \(x = 2\) into \(h(x)\): \(h(2)=3(2)^2-12(2)+9=12 - 24 + 9=-3\). Vertex is \((2, -3)\).

Step4: Find y - intercept

Set \(x = 0\): \(h(0)=3(0)^2-12(0)+9 = 9\). Y - intercept is \((0, 9)\).

Step5: Find x - intercepts

Set \(h(x)=0\): \(3x^2-12x + 9 = 0\). Divide by 3: \(x^2-4x + 3 = 0\). Factor: \((x - 1)(x - 3)=0\). So \(x = 1\) or \(x = 3\). X - intercepts are \((1, 0)\) and \((3, 0)\).

Step6: Find other points

Let's take \(x = 4\): \(h(4)=3(16)-12(4)+9=48 - 48 + 9 = 9\). Point is \((4, 9)\). Let's take \(x=-1\): \(h(-1)=3(1)+12 + 9=24\). Point is \((-1, 24)\).

Step7: Sketch the graph

Plot the vertex \((2, -3)\), axis of symmetry \(x = 2\), intercepts \((0, 9)\), \((1, 0)\), \((3, 0)\) and other points \((4, 9)\), \((-1, 24)\). Draw a parabola opening upward through these points.

Answer:

The parabola \(h(x)=3x^2 - 12x + 9\) opens upward, has axis of symmetry \(x = 2\), vertex \((2, -3)\), y - intercept \((0, 9)\), x - intercepts \((1, 0)\) and \((3, 0)\), and other points like \((4, 9)\) and \((-1, 24)\) (graph is sketched using these features).