Question

Plot points between and beyond the x-intercepts and the vertical asymptote. Evaluate the function.

Answer 1

Let's determine the values of f(x) based on the given x-values using the following function:

`\text{ f\lparen x\rparen = }\frac{\text{ x}^2\text{ + x - 42}}{\text{ x - 7}}`

At x = -9,

`\text{ f\lparen-9\rparen= }\frac{(-9)^2\text{ + \lparen-9\rparen- 42 }}{-9\text{ - 7}}\text{ = }\frac{81\text{ - 9 - 42}}{-9\text{ - 7}}\text{ = }(30)/(-16)\text{ = -}(15)/(8)\text{ \lparen Simplified\rparen}`

At x = -8,

`\text{ f\lparen-8\rparen = }\frac{(-8)^2\text{ + \lparen-8\rparen - 42}}{-8\text{ - 7}}\text{ = }\frac{64\text{ - 8 - 42}}{-8\text{ - 7}}\text{ = }(14)/(-15)\text{ = -}(14)/(15)`

At x = -1,

`\text{ f\lparen-1\rparen = }\frac{(-1)^2\text{ +}(-1)\text{ - 42}}{-1\text{ - 7}}\text{ = }\frac{1\text{ - 1 - 42}}{-1\text{ - 7}}\text{ = }(-42)/(-8)\text{ = }(21)/(4)\text{ \lparen Simplified\rparen}`

At x = 15/2,

`\text{ f\lparen}(15)/(2))\text{ = }\frac{((15)/(2))^2\text{ + \lparen}(15)/(2))\text{ - 42}}{(15)/(2)\text{ - 7}}\text{ = }\frac{(225)/(4)+(30)/(4)\text{ - }(168)/(4)}{(30)/(4)\text{ - }(28)/(4)}\text{ = }\frac{225\text{ + 30 - 168}}{30\text{ - 28}}\text{ = }(87)/(2)`

At x = 10,

`\text{ f\lparen10\rparen = }\frac{(10)^2\text{ + \lparen10\rparen - 42}}{\text{ 10 - 7}}\text{ = }\frac{\text{ 100 + 10 - 42}}{10\text{ - 7}}\text{ = }(68)/(3)`

In Summary,

At x = -9, f(-9) = -15/8

At x = -8, f(-8) = -14/15

At x = -1, f(-1) = 21/4

At x = 15/2, f(15/2) = 87/2

At x = 10, f(10) = 68/3