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 quadratic function is in the form \( h(x) = ax^2 + bx + c \), here \( a = 3 \). Since \( a>0 \), the parabola opens upward.

Step2: Find Vertex and Axis of Symmetry

The x - coordinate of the vertex (and the equation of the axis of symmetry) is given by \( x=-\frac{b}{2a} \). For \( h(x)=3x^{2}-12x + 9 \), \( a = 3 \), \( b=- 12 \). So \( x=-\frac{-12}{2\times3}=\frac{12}{6} = 2 \).
To find the y - coordinate of the vertex, substitute \( x = 2 \) into \( h(x) \): \( h(2)=3\times(2)^{2}-12\times2 + 9=3\times4-24 + 9=12-24 + 9=-3 \). So the vertex is \( (2,-3) \) and the axis of symmetry is \( x = 2 \).

Step3: Find y - intercept

The y - intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into \( h(x) \): \( h(0)=3\times(0)^{2}-12\times0 + 9=9 \). So the y - intercept is \( (0,9) \).

Step4: Find x - intercepts

Set \( h(x)=0 \), so \( 3x^{2}-12x + 9 = 0 \). Divide through by 3: \( x^{2}-4x + 3=0 \). Factor the quadratic: \( (x - 1)(x - 3)=0 \). Set each factor equal to zero: \( x-1=0\Rightarrow x = 1 \); \( x - 3=0\Rightarrow x = 3 \). So the x - intercepts are \( (1,0) \) and \( (3,0) \).

Step5: Find Two Other Points

Let's choose \( x=4 \). Then \( h(4)=3\times(4)^{2}-12\times4 + 9=3\times16-48 + 9=48-48 + 9=9 \). So the point is \( (4,9) \).
Let's choose \( x = 1.5 \). \( h(1.5)=3\times(1.5)^{2}-12\times1.5 + 9=3\times2.25-18 + 9=6.75-18 + 9=-2.25 \). So the point is \( (1.5,-2.25) \) (or we can use the symmetry. Since the axis of symmetry is \( x = 2 \), the point symmetric to \( (0,9) \) about \( x = 2 \) is \( (4,9) \) as we found, and the point symmetric to \( (1,0) \) about \( x = 2 \) is \( (3,0) \), but we can also calculate a non - symmetric point like \( (1.5,-2.25) \)).

To sketch the graph:

• Draw the coordinate plane.
• Plot the vertex \( (2,-3) \), the axis of symmetry \( x = 2 \) (a vertical dashed line).
• Plot the y - intercept \( (0,9) \), x - intercepts \( (1,0) \), \( (3,0) \) and the additional point \( (4,9) \).
• Since the parabola opens upward, draw a smooth curve through these points, symmetric about \( x = 2 \).

Answer:

• Direction of opening: Upward
• Vertex: \( (2,-3) \)
• Axis of symmetry: \( x = 2 \)
• y - intercept: \( (0,9) \)
• x - intercepts: \( (1,0) \), \( (3,0) \)
• Additional points: \( (4,9) \), \( (1.5,-2.25) \) (and the graph is a parabola opening upward with the above - mentioned features)