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Los puntos de intersección de los gráficos de f(x)=x2+6x−7 y g(x)=4x−10

User Javier Rey
by
7.9k points

1 Answer

3 votes

Answer:

Those graphs do not intersect.

Estes gráficos no se intersecciónan

Explanation:

The intersection points are x for which:


f(x) = g(x)

In this question:


f(x) = x^(2) + 6x - 7


g(x) = 4x - 10

So


x^(2) + 6x - 7 = 4x - 10


x^(2) + 2x + 3 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:


ax^(2) + bx + c, a\\eq0.

This polynomial has roots
x_(1), x_(2) such that
ax^(2) + bx + c = a(x - x_(1))*(x - x_(2)), given by the following formulas:


x_(1) = (-b + √(\bigtriangleup))/(2*a)


x_(2) = (-b - √(\bigtriangleup))/(2*a)


\bigtriangleup = b^(2) - 4ac

In this question:


x^(2) + 2x + 3 = 0

So
a = 1, b = 2, c = 3


\bigtriangleup = b^(2) - 4ac = 2^(2) - 4*1*3 = -8

Sincce
\bigtriangleup is negative, there are no solutions, which means that those graphs do not intersect.

User Cavalcantelucas
by
8.0k points
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