题目

Problem

Figure 2 shows a sketch of part of the graph with equation $y=f(x)$ , where

f(x)=\frac12|2x+7|-10

(a) State the coordinates of the vertex, $V$ , of the graph.

(2)

(b) Solve, using algebra,

\frac12|2x+7|-10\ge \frac13x+1

(4)

y=|f(x)|

stating the coordinates of the local maximum point and each local minimum point.

(4)

题目中文翻译

图 2 给出了部分图像 $y=f(x)$ 的示意图，其中

f(x)=\frac12|2x+7|-10

(a) 写出该图像顶点 $V$ 的坐标。

(b) 用代数方法解不等式

\frac12|2x+7|-10\ge \frac13x+1

y=|f(x)|

的图像，并写出局部极大点以及各局部极小点的坐标。

解答

The vertex occurs when the expression inside the modulus is zero:

2x+7=0

x=-\frac72

At this point,

f(x)=\frac12\cdot0-10=-10

Therefore

\boxed{V=\left(-\frac72,-10\right)}

We need to solve

\frac12|2x+7|-10\ge \frac13x+1

Add $10$ to both sides:

\frac12|2x+7|\ge \frac13x+11

Multiply by $2$ :

|2x+7|\ge \frac23x+22

The critical values occur when

2x+7=\frac23x+22

-(2x+7)=\frac23x+22

First,

2x+7=\frac23x+22

\frac43x=15

and hence

x=\frac{45}{4}

Second,

-2x-7=\frac23x+22

-\frac83x=29

and hence

x=-\frac{87}{8}

Because the modulus graph lies above the line outside the two intersection points, the solution is

\boxed{x\le -\frac{87}{8}\quad \text{or}\quad x\ge \frac{45}{4}}

For

y=|f(x)|

the part of $y=f(x)$ below the $x$ -axis is reflected in the $x$ -axis.

The vertex

\left(-\frac72,-10\right)

therefore becomes a local maximum point:

\left(-\frac72,10\right)

The local minimum points occur where

f(x)=0

\frac12|2x+7|-10=0

Hence

|2x+7|=20

Therefore

2x+7=20

2x+7=-20

So the local minimum points are

\left(\frac{13}{2},0\right) \qquad\text{and}\qquad \left(-\frac{27}{2},0\right)

Thus the local maximum point is

\boxed{\left(-\frac72,10\right)}

and the local minimum points are

\boxed{\left(-\frac{27}{2},0\right),\ \left(\frac{13}{2},0\right)}