Solve Lesson 16 Page 48 Math 10 SBT – Kite>

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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 \(\ left( {\frac{5}{3}; + \infty } \right)\)

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