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

Answer:

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

The largest xx-intercept is $x = 6$

The yy-intercept is $y = 30$.

Step-by-step explanation:

Given a quadratic function in the following format:

$f(x) = ax^{2} + bx + c = 0, a \neq 0$

The x-values of the x-intercepts are $x_{1}, x_{2}$, given by the following formulas.

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

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, $a = 1, b = -1, c = -30$

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

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-1) + \sqrt{121}}{2(1)} = \frac{12}{2} = 6$

$x_{2} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-1) - \sqrt{121}}{2(1)} = \frac{-10}{2} = -5$

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 $y = 30$.