Explanation:
Given the function:
[tex]y=2x^2-12x+15[/tex]
First, we find the vertex of the parabola.
Vertex
The equation of the axis of symmetry is calculated using the formula:
[tex]x=-\frac{b}{2a}[/tex]
From the function: a=2, b=-12
[tex]\implies x=-\frac{-12}{2(2)}=\frac{12}{4}=3[/tex]
Substitute x=3 into y to find the y-coordinate at the vertex.
[tex]\begin{gathered} y=2x^2-12x+15 \\ =2(3)^2-12(3)+15 \\ =18-36+15 \\ =-3 \end{gathered}[/tex]
The vertex is at (3, -3).
Two points to the left of the vertex
When x=2
[tex]\begin{gathered} y=2x^2-12x+15 \\ =2(2)^2-12(2)+15=8-24+15=-1 \\ \implies(2,-1) \end{gathered}[/tex]
When x=1
[tex]\begin{gathered} y=2x^2-12x+15 \\ =2(1)^2-12(1)+15=2-12+15=5 \\ \implies(1,5) \end{gathered}[/tex]
Two points to the right of the vertex
When x=4
[tex]\begin{gathered} y=2x^2-12x+15 \\ =2(4)^2-12(4)+15=32-48+15=-1 \\ \implies(4,-1) \end{gathered}[/tex]
When x=5
[tex]\begin{gathered} y=2x^2-12x+15 \\ =2(5)^2-12(5)+15=50-60+15=5 \\ \implies(5,5) \end{gathered}[/tex]
Answer:
Plot these points on the graph: (3, -3), (2,-1), (1,5), (4,-1), and (5,5).
