We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
51
2
avatar

What real value of t produces the smallest value of the quadratic t^2 -9t - 36?

 Oct 19, 2019
 #1
avatar
0

Solve for t:
t^2 - 9 t - 36 = 0

The left hand side factors into a product with two terms:
(t - 12) (t + 3) = 0

Split into two equations:
t - 12 = 0 or t + 3 = 0

Add 12 to both sides:
t = 12 or t + 3 = 0

Subtract 3 from both sides:
 
t = 12       or        t = -3

 Oct 19, 2019
 #2
avatar+196 
0

To find the smallest value, you need to find the vertex of the porabola. The vertex is the point that is the lowest on the porabola (or highest if a is negative). The x value produces the lowest (or highest) value of the porabola, so we only need to find one half of the point.

To find the x value, plug the coefficients into \({-b \over 2a}\)

\({-(-9) \over 2(1)}\)

=9/2

=4.5 will produce the lowest value of the porabola

 Oct 20, 2019

5 Online Users

avatar
avatar