sketch the graph of the quadratic function $h(x)=3x^{2}-12x + 9$ on the coordinate plane, ensuring to…

Explanation:

Step1: Determine the direction of opening

For a quadratic function $y = ax^{2}+bx + c$, here $a = 3>0$, so the parabola opens upward.

Step2: Find the axis of symmetry

The formula for the axis of symmetry of a quadratic function $y=ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Given $a = 3$ and $b=-12$, then $x =-\frac{-12}{2\times3}=2$.

Step3: Find the vertex

Substitute $x = 2$ into $h(x)=3x^{2}-12x + 9$. $h(2)=3\times2^{2}-12\times2 + 9=3\times4-24 + 9=12-24 + 9=-3$. So the vertex is $(2,-3)$.

Step4: Find the y - intercept

Set $x = 0$ in $h(x)$. $h(0)=3\times0^{2}-12\times0 + 9=9$. So the y - intercept is $(0,9)$.

Step5: Find the x - intercepts

Set $h(x)=0$, so $3x^{2}-12x + 9 = 0$. Divide through by 3: $x^{2}-4x + 3=0$. Factor it as $(x - 1)(x - 3)=0$. Then $x=1$ or $x = 3$. The x - intercepts are $(1,0)$ and $(3,0)$.

Step6: Find two other points

Let $x=0.5$, $h(0.5)=3\times(0.5)^{2}-12\times0.5 + 9=3\times0.25-6 + 9=0.75+3=3.75$. So the point is $(0.5,3.75)$. Let $x=2.5$, $h(2.5)=3\times(2.5)^{2}-12\times2.5 + 9=3\times6.25-30 + 9=18.75-30 + 9=-2.25$. So the point is $(2.5,-2.25)$.

Answer:

• Direction: Opens upward.
• Axis of symmetry: $x = 2$.
• Vertex: $(2,-3)$.
• y - intercept: $(0,9)$.
• x - intercepts: $(1,0)$ and $(3,0)$.
• Other points: $(0.5,3.75)$ and $(2.5,-2.25)$. Sketch the parabola using these points.