Explanation:
Given the function:
![y=2x^2-12x+15](https://img.qammunity.org/2023/formulas/mathematics/college/s5dbnsnlxa3wcuu6yxeogvzq5g9khfyo1x.png)
First, we find the vertex of the parabola.
Vertex
The equation of the axis of symmetry is calculated using the formula:
![x=-(b)/(2a)](https://img.qammunity.org/2023/formulas/mathematics/college/7gr846x3106wifbv8ib3mo7x3lghpti0f2.png)
From the function: a=2, b=-12
![\implies x=-(-12)/(2(2))=(12)/(4)=3](https://img.qammunity.org/2023/formulas/mathematics/college/4qedhhbektjk60ll9x020r75u9pva60nm4.png)
Substitute x=3 into y to find the y-coordinate at the vertex.
![\begin{gathered} y=2x^2-12x+15 \\ =2(3)^2-12(3)+15 \\ =18-36+15 \\ =-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ra3vq330s2khkakpjcu7tsvqihiwrl74lq.png)
The vertex is at (3, -3).
Two points to the left of the vertex
When x=2
![\begin{gathered} y=2x^2-12x+15 \\ =2(2)^2-12(2)+15=8-24+15=-1 \\ \implies(2,-1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/45i1b0y6atpngi9pwgsaxc6zp8hzlvfm5z.png)
When x=1
![\begin{gathered} y=2x^2-12x+15 \\ =2(1)^2-12(1)+15=2-12+15=5 \\ \implies(1,5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c847u8kq7z4vskpm522m3xoilb1t3is042.png)
Two points to the right of the vertex
When x=4
![\begin{gathered} y=2x^2-12x+15 \\ =2(4)^2-12(4)+15=32-48+15=-1 \\ \implies(4,-1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/671ka8sf6lx3uqwx7uj4ztwx6lomzqla06.png)
When x=5
![\begin{gathered} y=2x^2-12x+15 \\ =2(5)^2-12(5)+15=50-60+15=5 \\ \implies(5,5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xs4qt2rosz2ta4a2ncsms334rrt0t31swk.png)
Answer:
Plot these points on the graph: (3, -3), (2,-1), (1,5), (4,-1), and (5,5).