Question

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

Answer 1

Answer:

Those graphs do not intersect.

Estes gráficos no se intersecciónan

Step-by-step 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:

$ax^{2} + bx + c, a\neq0$.

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:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

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.