Answer:
[tex]\textsf{(a)}\quad (-1, 5)[/tex]
[tex]\textsf{(b)} \quad y=\dfrac{1}{4}(x-3)^2+1[/tex]
Step-by-step explanation:
Part (a)
If the vertex of the quadratic function is (3, 1) then the function's axis of symmetry is x=3.
Symmetric points on the curve of a quadratic function have the same y-value and are the same distance (but in opposite horizontal directions) from the axis of symmetry.
The x-value of point (7, 5) is 4 units to the right from the axis of symmetry.
Therefore, the other point that is symmetrical to point (7, 5) has an x-value that is 4 units to the left of the axis of symmetry.
Therefore, another point on the function's graph is:
⇒ (3-4, 5) = (-1, 5)
Part (b)
[tex]\boxed{\begin{minipage}{5.6 cm}\underline{Vertex form of a quadratic equation}\\\\$y=a(x-h)^2+k$\\\\where:\\ \phantom{ww}$\bullet$ $(h,k)$ is the vertex. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}[/tex]
Given information:
- Vertex = (3, 1)
- Point on the curve = (7, 5)
Therefore:
Substitute the values into the formula and solve for a:
[tex]\implies 5=a(7-3)^2+1[/tex]
[tex]\implies 5=a(4)^2+1[/tex]
[tex]\implies 5=16a+1[/tex]
[tex]\implies 4=16a[/tex]
[tex]\implies a=\dfrac{1}{4}[/tex]
Therefore, the equation of the quadratic function in vertex form is:
[tex]\boxed{y=\dfrac{1}{4}(x-3)^2+1}[/tex]
