Question

Consider the quadratic function f(x)=−x^2+x+30

Determine the following:

The smallest xx-intercept is x=
The largest xx-intercept is x=
The yy-intercept is y=

Answer 1

The smallest xx-intercept is
`x = -5`

The largest xx-intercept is
`x = 6`

The yy-intercept is
`y = 30`.

Explanation:

Given a quadratic function in the following format:

`f(x) = ax^(2) + bx + c = 0, a \\eq 0`

The x-values of the x-intercepts are
`x_(1), x_(2)`, given by the following formulas.

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

We have that:

`f(x) = -x^(2) + x + 30`

This is not in the format above. I will multiply by (-1), so we have:

`f(x) = x^(2) - x - 30`

So,
`a = 1, b = -1, c = -30`

`\bigtriangleup = b^(2) - 4ac = (-1)^(2) - 4*(1)*(-30) = 1 + 120 = 121`

`x_(1) = (-b + √(\bigtriangleup))/(2*a) = (-(-1) + √(121))/(2(1)) = (12)/(2) = 6`

`x_(2) = (-b + √(\bigtriangleup))/(2*a) = (-(-1) - √(121))/(2(1)) = (-10)/(2) = -5`

This means that

The smallest xx-intercept is
`x = -5`

The largest xx-intercept is
`x = 6`

The y-intercept is the value of f(x) when x = 0. So

`f(x)=−x^(2)+x+30`

`f(0) = 30`

The yy-intercept is
`y = 30`.