Question

Write a formula for a quadratic function with the indicated characteristics. Has horizontal intercepts at t = 3 and t = 4 and graph containing the point (1, -3)

Answer 1

Answer:

$\text{ y = -}\frac{1}{2}(t^2-7t+12)$

Explanation:

Here, we want to write the equation of the quadratic function

We have the general form as:

$\text{ y = ax}^2\text{ + bx + c}$

from the question, we have the horizontal intercepts at t= 3 and t =4

that means (t-3) is a factor and also (t-4) is another root of the equation

The product of these two is as follows:

$\text{ a(t-3)(t-4) = a(t}^2-7t\text{ + 12)}$

Finally, we need to find the value of a

We can do this by making a substitution

We substitute 1 for t and -3 on the other side of the equation

Mathematically, we have this as:

$\begin{gathered} \text{ a(1-7+12) = -3} \\ 6a\text{ = -3} \\ a\text{ = }\frac{-3}{6}\text{ = -}\frac{1}{2} \end{gathered}$

Thus, we have the equation as:

$\text{ y = -}\frac{1}{2}(t^2-7t+12)$