Question

we want to find the intersection points of the graphs given by the following system of equations:
\\( \left\\{ \

$$\begin{array} { l } { x - y = - 4 } \\\\ { y = 5 ( x + 1 ) ^ { 2 } - 3 } \\end{array}$$

\
ight. \\)
one of the intersection points is \\( ( - 2,2 ) \\).
find the other intersection point.
your answer must be exact.
related content
quadratic systems: a line and a parabola
quadratic systems: a line and a circle

Explanation:

Step1: Substitute \(y\) from the first equation into the second

From \(x - y=-4\), we have \(y=x + 4\). Substitute \(y=x + 4\) into \(y = 5(x + 1)^{2}-3\). So, \(x+4=5(x + 1)^{2}-3\).
Expand \(5(x + 1)^{2}-3\): \(5(x^{2}+2x + 1)-3=5x^{2}+10x+5 - 3=5x^{2}+10x + 2\).
The equation becomes \(x + 4=5x^{2}+10x+2\).
Rearrange to get a quadratic equation: \(5x^{2}+9x - 2=0\).

Step2: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 5\), \(b = 9\), \(c=-2\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(9)^{2}-4\times5\times(-2)=81 + 40=121\).
Then \(x=\frac{-9\pm\sqrt{121}}{2\times5}=\frac{-9\pm11}{10}\).
We know one solution: when \(x=\frac{-9 - 11}{10}=\frac{-20}{10}=-2\) (already given).
For the other solution, \(x=\frac{-9 + 11}{10}=\frac{2}{10}=\frac{1}{5}\).

Step3: Find the corresponding \(y\) - value

Substitute \(x=\frac{1}{5}\) into \(y=x + 4\). So \(y=\frac{1}{5}+4=\frac{1+20}{5}=\frac{21}{5}\).

Answer:

\((\frac{1}{5},\frac{21}{5})\)