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Evaluate the function f(x) = x2 + 3x -2 for each replacement value of x.
a. -3
b. 0
c. 2
d. 3
e. -2/3
f. 0.005


Sagot :

Step-by-step explanation:

a. x = -3

[tex]f( - 3) = { (- 3)}^{2} + 3( - 3) - 2[/tex]

[tex]f( - 3) = 9 - 9 - 2[/tex]

[tex]f( - 3 ) = - 2[/tex]

b. x = 0

[tex]f(0) = {(0)}^{2} + 3(0) - 2[/tex]

[tex]f(0) = - 2[/tex]

c. x = 2

[tex] f(2) = {(2)}^{2} + 3(2) - 2[/tex]

[tex]f(2) = 4 + 6 - 2[/tex]

[tex]f(2) = 10 - 2[/tex]

[tex]f(2) = 8[/tex]

d. x = 3

[tex]f(3) = {(3)}^{2} + 3(3) - 2[/tex]

[tex]f(3) = 9 + 9 - 2[/tex]

[tex]f(3) = 9 - 2[/tex]

[tex]f(3 ) = 7[/tex]

e. x = - 2/3

[tex]f( - \frac{2}{3}) = {( - \frac{2}{3}) }^{2} + 3( - \frac{2}{3}) - 2 [/tex]

[tex]f( - \frac{2}{3}) = \frac{4}{9} - 2 - 2[/tex]

[tex]f( - \frac{2}{3} ) = \frac{4}{9} - 4[/tex]

[tex]f( - \frac{2}{3}) = \frac{4}{9} - \frac{4}{1} = \frac{4 - 36}{9} = - \frac{32}{9} [/tex]

[tex]f( - \frac{2}{3 } ) = - \frac{32}{9} [/tex]

f. x = 0.005

[tex]f(0.005) = {(0.005)}^{2} + 3(0.005) - 2[/tex]

[tex]f(0.005) = 0.000025 + 0.015 - 2[/tex]

f(0.005) = -1.984975