Question

1 mode → par (parametric)

1. x1t = 5cos(t), y1t = 5sin(t).

3 graph → sketch circle.
part 6: exit ticket (5 minutes)
graph y=-x² + 6x - 5.

1. enter in y=.
2. graph

3 2nd → trace → 3 minimum.
vertex =
sketch graph here

Explanation:

Step1: Recall vertex - formula for parabola

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$. In the function $y=-x^{2}+6x - 5$, $a=-1$ and $b = 6$.
$x=-\frac{6}{2\times(-1)}=3$

Step2: Find the $y$ - coordinate of the vertex

Substitute $x = 3$ into the function $y=-x^{2}+6x - 5$.
$y=-(3)^{2}+6\times3 - 5=-9 + 18-5=4$

Answer:

Vertex: $(3,4)$