Mth 163 040129
What are coordinates of the vertex of the quadratic function y(t) = .02 t^2 – 3 t + 80?
The vertex is at t = -b / (2a) = -(-3) / (2 * .02) = ...
To find the y coordinate of the vertex substitute this value of t into the function.
Having found the coordinates vertex, how do you quickly (without actually evaluating the function) find the coordinates of the points that lie on the graph 1 unit to the right and 1 unit to the left of the vertex? What are the coordinates of these points for the function y(t) = .02 t^2 – 3 t + 80?
If we move 1 unit right or left of the vertex the y coordinate changes by a = .02.
If the vertex is at (74, -32.5) then we get (74, -32.48) and (76, -32.48).
What are the coordinates of the points at which the graph of the function y(t) = .02 t^2 – 3 t + 80 crosses the t axis?
Plug a = .02, b = 3 and c =80 into the quadratic formula. If the vertex is below the t axis and the parabola opens upward, which are both indicated by the preceding, there will be two zeros. This will happen because the discriminant b^2 - 4 a c will be positive, giving us two different values for the +- of the formula. These two values will be at equal distances from the axis of symmetry t = -b / (2a) = 75.
Quickly make a table for the y = x^2 parabola for x values from -2 to 2, and quickly sketch the graph of this parabola.
You should be able to do this in a minute or less, awake or asleep, rested or tarred.
Quickly show how you can use the graph of y = x^2 and the concept of stretching to obtain the graph of y = 1 / 2 * x^2, without actually making a table for this function.
Note that points at different distances from the x axis will move different distances, since half of a bigger number is greater than half of a smaller number.
All points move to half their distance from the x axis.
To construct a graph of the quadratic y(t) = .02 t^2 – 3 t + 80 by starting with a graph of y = t^2, what series of stretches and shifts would you use?
Stretch by factor .02 to get the graph of y = .02 t^2.
Horizontally shift 75 units to get the graph of y = .02 ( t - 75)^2.
Vertically shift -32.5 units to get the graph of y = .02 ( t - 75)^2 - 32.5.
If f(t) = .02 t^2 – 3 t + 80 then give expressions for the following:
f(10) = .02 * 10^2 - 3 * 10 + 80 = ... = 52 (?)
f(a+b) = .02 ( a+b)^2 - 3(a+b) + 80 = ... etc. (no F***; use Distributive Law ).
We get
.02 a^2 + .04 a b + .02 b^2 - 3 a - 3 b + 80.