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Question: How do you find the point of intersection of\[f(x) = x\] and\[g(x) = - x - 2?\]...

How do you find the point of intersection off(x)=xf(x) = x andg(x)=x2?g(x) = - x - 2?

Explanation

Solution

The given question describes the operation of addition/ subtraction/ multiplication/ division. We need to know the condition which is used to find the intersection points of two lines. In this question, we have to find the valuexxby using a suitable formula. Also, we need to know the multiplication between a positive number and a negative number.

Complete step by step solution:
To solve the given question, we have to find the intersection point of f(x)=xf(x) = xandg(x)=x2g(x) = - x - 2.

The condition to find the intersection point between the given equations is,
f(x)=g(x)(1)f(x) = g(x) \to \left( 1 \right)

We know that,f(x)=xf(x) = x g(x)=x2g(x) = - x - 2

Let’s substitute the above values in the equation(1)\left( 1 \right) we get,

f(x)=g(x)f(x) = g(x)
x=x2x = - x - 2

To make easy calculations we have to addxx LHS and RHS to the equation. So, we get
x+x=x2+xx + x = - x - 2 + x

On the right-hand side of the equation, we havex - xandxx, both are cancelled by each other.

So, we get

x+x=2(2)x + x = 2 \to \left( 2 \right)
We know that, x+xx + x
\downarrow
2x2x

So, the equation(2)\left( 2 \right)becomes
2x=22x = - 2

Let’s move the term22from the left side to right side of the equation in the2x2xform. So, we get

x=22x = \dfrac{{ - 2}}{2}
x=1x = - 1

So, the final answer is,
x=1x = - 1Is the intersection point of,f(x)=xf(x) = xand g(x)=x2g(x) = - x - 2.

Note: The intersection point is the point at where the given two line values are equal(i.ef(x)=g(x))(i.ef(x) = g(x)).

To make an easy calculation we can add/ subtract/ multiply/ divide any term with the equation into both sides. We shouldn’t multiply zero with the equation, otherwise, we didn’t get the final answer. When multiplying the different sign numbers we would remind the following things,

()×(+)=()\left( - \right) \times \left( + \right) = \left( - \right)

()×()=(+)\left( - \right) \times \left( - \right) = \left( + \right)

(+)×(+)=(+)\left( + \right) \times \left( + \right) = \left( + \right)