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Masja [62]
3 years ago
11

Which graph correctly solves the system of equations below? y = − x2 + 1 y = x2 − 4 (6 points)

Mathematics
2 answers:
labwork [276]3 years ago
4 0

Answer:

<em>Solve for the first variable in one of the equations, then substitute the result into the other equation.</em>

<em>Solve for the first variable in one of the equations, then substitute the result into the other equation.Point Form:</em>

<em>(√102,−32),(−√102,−32)</em>

<em>Equation Form:</em>

<em>x</em><em>=√10</em><em>/</em><em>2,y</em><em> </em><em>=</em><em> </em><em>−3</em><em>/</em><em>2</em>

<em>x</em><em>=−√10</em><em>/</em><em>2,</em><em>y</em><em> </em><em>=</em><em> </em><em>−3</em><em>/</em><em>2</em>

IgorC [24]3 years ago
3 0

The equations are  

y = -x^{2} - 1

y = 2x^{2} -4

The graphs of the solutions (x, y) of these equations are 2 parabolas, since the right hand side expressions are polynomials of degree 2.

The solution/s  of the system are the x-coordinates of the point/s of intersection of the parabolas.

The solutions of the first equation form a parabola looking downwards (since the coefficient of x^2 is -), and the second, a parabola opening upwards (since the coefficient of x^2 is +).

We can draw both parabolas, but to find the solution we still need to solve the system algebraically.

The algebraic solution of the system is:

-x^{2} -1 =2x^{2} - 4

3x^{2} -3 =0

3(x^{2} -1) = 0

x^{2} -1 =0     , so

the solutions are x=-1 and x=1.  

If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.

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