Difference between revisions of "8A F11 Q5"
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! Step 1: | ! Step 1: | ||
|- | |- | ||
| − | | | + | |First we replace the inequalities with equality. So <math>y = \vert x\vert + 1</math>, and <math>x^2 + y^2 = 9</math>. |
| + | |- | ||
| + | |Now we graph both functions. | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Step 2: | ||
| + | |- | ||
| + | |Now that we have graphed both functions we need to know which region to shade with respect to each graph. | ||
| + | |- | ||
| + | |To do this we pick a point an equation and a point not on the graph of that equation. We then check if the | ||
| + | |- | ||
| + | |point satisfies the inequality or not. For both equations we will pick the origin. | ||
| + | |- | ||
| + | |<math>y < \vert x\vert + 1:</math> Plugging in the origin we get, <math>0 < \vert 0\vert + 1 = 1</math>. Since the inequality is satisfied shade the side of | ||
| + | |- | ||
| + | |<math>y < \vert x\vert + 1</math> that includes the origin. We make the graph of <math>y < \vert x\vert + 1</math>, since the inequality is strict. | ||
| + | |- | ||
| + | |<math>x^2 + y^2 \le 9:</math> <math>(0)^2 +(0)^2 = 0 \le 9</math>. Once again the inequality is satisfied. So we shade the inside of the circle. | ||
| + | |- | ||
| + | |We also shade the boundary of the circle since the inequality is <math>\le</math> | ||
| + | |} | ||
| + | |||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Step 3: | ||
| + | |- | ||
| + | |The final solution is the portion of the graph that is shaded by both inequalities | ||
|} | |} | ||
Revision as of 19:23, 22 March 2015
Question: Graph the system of inequalities
| Foundations |
|---|
| 1) What do the graphs of , and look like? |
| 2) Each graph splits the plane into two regions. Which one do you want to shade? |
| Answer: |
| 1) The first graph looks like a V with the vertex at (0, 1), the latter is a circle centered at the origin with radius 3. |
| 2) Since the Y-value must be less than , shade below the V. For the circle shde the inside. |
Solution:
| Step 1: |
|---|
| First we replace the inequalities with equality. So , and . |
| Now we graph both functions. |
| Step 2: |
|---|
| Now that we have graphed both functions we need to know which region to shade with respect to each graph. |
| To do this we pick a point an equation and a point not on the graph of that equation. We then check if the |
| point satisfies the inequality or not. For both equations we will pick the origin. |
| Plugging in the origin we get, . Since the inequality is satisfied shade the side of |
| that includes the origin. We make the graph of , since the inequality is strict. |
| . Once again the inequality is satisfied. So we shade the inside of the circle. |
| We also shade the boundary of the circle since the inequality is |
| Step 3: |
|---|
| The final solution is the portion of the graph that is shaded by both inequalities |