Given Function, find domain and description of graph $y = f(x)$

Question

I am studying for Graduate Record Exam.

The following question is difficult.

Given the domain and description of $f(x) = 5 - (x + 20)^2$, including its shape, and the $x$ and $y$-intercepts

To find the $y$-intercept, I make $x = 0$

$f(x) = 5 - (x + 20)^2$

$f(x) = 5 - (20)^2$

$f(x) = 5 - 400$

$f(x) = -395$

Now, I need to find $x$-intercept, so I set $y$ to $0$

$5 - (x + 20)^2 = 0$

$5 - (x^2 + 40x + 400) = 0$

becomes the quadratic equation

$5 - (x^2 + 40x + 400) = 0$

$5 - x^2 - 40x - 400 = 0$

$-x^2 - 40x - 395 = 0$

and $x$-intercepts are $-20 \pm \sqrt{5}$

However, the answer at the back of book says the $x$-intercepts are $-20 \pm \sqrt{5}$

How did I get my math wrong?

Answer 1

Your discriminant should be $ \ \sqrt{(40)^2 \ - \ 4 \ \cdot \ 1 \ \cdot \ 395} \ = \ \sqrt{1600 \ - \ 1580 \ } \ = \ \sqrt{20} \ = \ \ 2 \sqrt{5} \ $ . Then remember to divide that by $ \ 2a \ = \ 2 \ \cdot \ 1 \ $ , as you did with the $ \ -40 \ $ .

Answer 2

1) $-x^2 - 40x - 395 = 0 \iff x^2 + 40x + 395=0$

$x = \frac{40 \pm \sqrt{40^2 - 4*395}}{2} = 20 \pm \frac{\sqrt{1600 - 1580}}{2}=20 \pm \frac{\sqrt{20}}{2}= 20 \pm \frac{2\sqrt{5}}{2} = 20 \pm \sqrt 5$

I made an error doing this too. I think maybe you did $\frac{\sqrt{20}}{2} = \frac{\sqrt{4*5}}{2}= \frac{4\sqrt{5}}{2}$ which is, of course, careless. (but I did it anyway...:(