Find the intersections, then the area

The intersections of $y=x^2$ and $y=x+12$ come from $x^2=x+12$, that is $x^2-x-12=(x+3)(x-4)=0$, so $x=-3,\ 4$.

Between the intersections the line lies above the parabola, so subtract the parabola from the line. The smaller $x$-coordinate is the lower limit and the larger one is the upper limit.

Examples

1. 01What number goes in □?$\displaystyle \begin{gathered} \displaystyle y=x^2, \\[6pt] \displaystyle y=25 \;\Rightarrow\; x=\pm\square \end{gathered}$AnswerHide

   Answer$\displaystyle 5$

   Steps

   1. $\displaystyle x^2 = 25$$\displaystyle \Rightarrow x = \pm 5$
2. 02What are the x-coordinates of the intersections of the two curves?$\displaystyle y=x^2,\quad y=x+12$AnswerHide

   Answer$\displaystyle x=-3,\ 4$

   Steps

   1. $\displaystyle x^2 - (x+12)$$\displaystyle = (x+3)(x-4)$
   2. A product is 0, so one factor is 0
   3. $\displaystyle (x+3)(x-4) = 0$$\displaystyle \Rightarrow x = -3$$\displaystyle (x+3)(x-4) = 0$$\displaystyle \Rightarrow x = 4$
3. 03What are the x-coordinates of the intersections of the two curves?$\displaystyle y=x^2,\quad y=-x+12$AnswerHide

   Answer$\displaystyle x=-4,\ 3$

   Steps

   1. $\displaystyle x^2 - (-x+12)$$\displaystyle = (x+4)(x-3)$
   2. A product is 0, so one factor is 0
   3. $\displaystyle (x+4)(x-3) = 0$$\displaystyle \Rightarrow x = -4$$\displaystyle (x+4)(x-3) = 0$$\displaystyle \Rightarrow x = 3$
4. 04What are the x-coordinates of the intersections of the two curves?$\displaystyle y=x^2,\quad y=-x+2$AnswerHide

   Answer$\displaystyle x=-2,\ 1$

   Steps

   1. $\displaystyle x^2 - (-x+2)$$\displaystyle = (x+2)(x-1)$
   2. A product is 0, so one factor is 0
   3. $\displaystyle (x+2)(x-1) = 0$$\displaystyle \Rightarrow x = -2$$\displaystyle (x+2)(x-1) = 0$$\displaystyle \Rightarrow x = 1$
5. 05What are the x-coordinates of the intersections of the two curves?$\displaystyle y=x^2,\quad y=2x$AnswerHide

   Answer$\displaystyle x=0,\ 2$

   Steps

   1. $\displaystyle x^2 - 2x$$\displaystyle = x(x-2)$
   2. A product is 0, so one factor is 0
   3. $\displaystyle x(x-2) = 0$$\displaystyle \Rightarrow x = 0$$\displaystyle x(x-2) = 0$$\displaystyle \Rightarrow x = 2$
6. 06What are the x-coordinates of the intersections of the two curves?$\displaystyle y=x^2,\quad y=x+2$AnswerHide

   Answer$\displaystyle x=-1,\ 2$

   Steps

   1. $\displaystyle x^2 - (x+2)$$\displaystyle = (x+1)(x-2)$
   2. A product is 0, so one factor is 0
   3. $\displaystyle (x+1)(x-2) = 0$$\displaystyle \Rightarrow x = -1$$\displaystyle (x+1)(x-2) = 0$$\displaystyle \Rightarrow x = 2$

FAQ

Q1Is $x=5$ the only intersection of $y=x^2$ and $y=25$?AnswerHide

A

$x^2=25$ has two solutions, $x=\pm 5$. The graphs meet at $x=-5$ and $x=5$, and the enclosed region spans $-5\le x\le 5$.

Q2Can I divide both sides of $x^2=3x$ by $x$?AnswerHide

A

Dividing by $x$ is allowed only when $x\ne 0$, so the solution $x=0$ is lost. Factor instead: $x^2-3x=x(x-3)=0$ gives both $x=0$ and $x=3$.