We have the equation:
[tex]y=\frac{1}{4}(x-5)^2-3[/tex]
This is a parabola expressed in vertex form, where the vertex is (h,k) = (5,-3).
We have to graph the parabola. To do that we need another point, as we already know the vertex and, therefore, the axis of symmetry (x = 5).
We can find another point by giving a value to x and calculating y.
For example, with x = 3 we get:
[tex]\begin{gathered} y(3)=\frac{1}{4}(3-5)^2-3 \\ y(3)=\frac{1}{4}(-2)^2-3 \\ y(3)=\frac{1}{4}\cdot4-3 \\ y(3)=1-3 \\ y(3)=-2 \end{gathered}[/tex]
The point that belongs to the parabola when x = 3 is (3, -2).
Then, we can graph the two points and draw the parabola as:
Because of the symmetry at x = 5, we also know that two units to the right, at x = 7, we will have the same value of y that we have for x = 3.
With at least 3 points, we can graph a parabola.
The actual graph is:
If we want to add more precision to our graph, we can calculate more points that belong to the parabola.
For example, at the other side of the vertex, we can calculate the value of y for x = 6:
[tex]\begin{gathered} y(6)=\frac{1}{4}(6-5)^2-3 \\ y(6)=\frac{1}{4}(1)^2-3^{} \\ y(6)=\frac{1}{4}-3 \\ y(6)=\frac{1}{4}-\frac{12}{4}^{} \\ y(6)=-\frac{11}{4}=-2.75 \end{gathered}[/tex]
We can add this to the plot as:
We have to aproximate the position of y as the grid only shows integers and y = -2.75.
Answer:
The points in the parabola are (5,-3), (3,-2) and (6,-2.75). We need at least 3 points to plot a parabola.
