## Solve Lesson 16 Page 48 Math 10 SBT – Kite>

Topic

State the range of covariance and inverse range of each of the following functions:

a) $$y = 4{x^2} + 6x – 5$$

b) $$y = – 3{x^2} + 10x – 4$$

Solution method – See details Given the function $$y = a{x^2} + bx + c$$

Step 1: Determine the coefficients a, b, c. Calculate $$\frac{{ – b}}{{2a}}$$

Step 2:

+ If $$a > 0$$

The function is covariant on $$(\frac{{ – b}}{{2a}}; + \infty )$$ and inverse on $$( – \infty ;\frac{{ – b}}{{2a) }})$$

+ If $$a < 0$$

The function is covariant on $$( – \infty ;\frac{{ – b}}{{2a}})$$ and inverse on $$(\frac{{ – b}}{{2a}}; \infty )$$

Detailed explanation

a) The function$$y = 4{x^2} + 6x – 5$$ has $$a = 4,b = 6,c = – 5 \Rightarrow \frac{{ – b}}{{2a}} = \frac{{ – 6}}{{2.4}} = – \frac{3}{4}$$

Since $$a = 4 > 0$$ the function is covariate on the interval $$\left( { – \frac{3}{4}; + \infty } \right)$$, inversely over the interval $$\ left( { – \infty ; – \frac{3}{4}} \right)$$

b) The function $$y = – 3{x^2} + 10x – 4$$ has $$a = – 3,b = 10,c = – 4 \Rightarrow \frac{{ – b}}{{2a }} = \frac{{ – 10}}{{2.\left( { – 3} \right)}} = \frac{5}{3}$$

Since $$a = – 3 < 0$$ the function is covariate on the interval $$\left( { – \infty ;\frac{5}{3}} \right)$$, inversely over the interval $$\ left( {\frac{5}{3}; + \infty } \right)$$