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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)

1 Answer

4 votes

Answer:


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

Step-by-step 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{ = }(-3)/(6)\text{ = -}(1)/(2) \end{gathered}

Thus, we have the equation as:


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

User OguzGelal
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