First, let's rearrange the given equation into something more recognizable. If we add 13 to both sides, we now have the polynomial $x^{2}-10x+13$ . We can now use the quadratic formula to solve.

Remember that the quadratic formula is

$\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

Substitute the numbers from the equation into the formula.

$\frac{-(-10)+/-\sqrt{(-10)^{2}-4(1)(13) } }{2(1)}$

Simplify:

$\frac{10+/-\sqrt{100-52} }{2}$

$\frac{10+/-\sqrt{48} }{2}$

Here, I'm going to assume that there was a mistype in option B because if we divide out the 2 we end up with $5+/-\sqrt{48}$ .

Hope this helps!

6 0

3 years ago

Find quadratic polynomial which best fits the function f(x)=e^x and x=0, in the sense that g(0)= f(0), and g'(0) = f'(0), and g"

Dmitriy789 [7]

Let $g(x)=ax^2+bx+c$ . Then

$\begin{cases}g(0)=c\\g'(0)=b\\g''(0)=2a\end{cases}$

Meanwhile, since $f(x)=e^x$ , you have $f(0)=f'(0)=f''(0)=e^0=1$ .

It follows that $a=\dfrac12$ , $b=1$ , and $c=1$ , so that the quadratic fit for $f(x)=e^x$ that satisfies the given points is

$g(x)=\dfrac12x^2+x+1$

Note that this is just the second-order Taylor polynomial for $e^x$ about $x=0$ .

3 0

4 years ago