Question

Write the equation for the parabola that has x− intercepts (−2−√2 ,0) and (−2+√2 ,0) and passes through the point(−1,1).

Answer 1

Answer:

f(x)= -x²-4x-2

Step-by-step explanation (this way is not the shortest one):

1. for intercept (-2-√2;0): a(-2-√2)²+b(-2-√2)+c=0

for intercept (-2+√2;0): a(-2+√2)²+b(-2+√2)+c=0

for the point (-1;1): a(-1)²+b(-1)+c=1

2. using the three record written above, it is possible to make up the system of equations and calculate 'a', 'b' and 'c':

$\left \{ \begin{array}{ccc}a-b+c=1\\a(-2- \sqrt2)^2+b(-2- \sqrt2)+c=0\\a(-2+ \sqrt2)^2+b(-2+ \sqrt2)+c=0\end{array}$

$\left \{ \begin{array}{ccc}a=-1\\b=-4 \\c=-2\end{array}$

3. the required equation is: -x²-4x-2