题目

Problem

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

f(x)=|3x+a|+a

and where $a$ is a positive constant.

The graph has a vertex at the point $P$ , as shown in Figure 2.

(a) Find, in terms of $a$ , the coordinates of $P$ .

(2)

(b) Sketch the graph with equation $y=g(x)$ , where

g(x)=|x+5a|

On your sketch, show the coordinates, in terms of $a$ , of each point where the graph cuts or meets the coordinate axes.

(2)

The graph with equation $y=g(x)$ intersects the graph with equation $y=f(x)$ at two points.

(5)

题目中文翻译

图 2 给出了方程 $y=f(x)$ 所表示图像的草图，其中

f(x)=|3x+a|+a

且 $a$ 为正常数。

图像的顶点为图中的点 $P$ 。

(a) 用 $a$ 表示点 $P$ 的坐标。

(b) 画出方程 $y=g(x)$ 的图像草图，其中

g(x)=|x+5a|

并在草图上标出图像与坐标轴相交或相切的各点坐标（用 $a$ 表示）。

已知方程 $y=g(x)$ 的图像与方程 $y=f(x)$ 的图像相交于两点。

解答

(a)

The vertex occurs when the expression inside the absolute value is zero:

3x+a=0

x=-\frac{a}{3}

At this point,

f(x)=|0|+a=a

Therefore

\boxed{P\left(-\frac{a}{3},a\right)}

(b)

For

g(x)=|x+5a|

the graph is a V shape.

The vertex occurs when

x+5a=0

x=-5a

Thus the graph meets the $x$ -axis at

(-5a,0)

For the $y$ -intercept, set $x=0$ :

g(0)=|5a|=5a

So the graph cuts the $y$ -axis at

(0,5a)

(c)

We need to solve

|x+5a|=|3x+a|+a

Since $a>0$ , split the graph into the relevant straight-line pieces.

Left intersection

For the left intersection, $x+5a$ is positive and $3x+a$ is negative, so

|x+5a|=x+5a

and

|3x+a|=-(3x+a)

Hence

x+5a=-3x-a+a

x+5a=-3x

Therefore

4x=-5a

x=-\frac{5a}{4}

Now find $y$ using $g(x)$ :

y=\left|-\frac{5a}{4}+5a\right| =\left|\frac{15a}{4}\right| =\frac{15a}{4}

So one intersection is

\left(-\frac{5a}{4},\frac{15a}{4}\right)

Right intersection

For the right intersection,

x+5a>0,\qquad 3x+a>0

|x+5a|=x+5a

and

|3x+a|=3x+a

The equation becomes

x+5a=3x+a+a

x+5a=3x+2a

Hence

3a=2x

and therefore

x=\frac{3a}{2}

Now find $y$ :

y=\frac{3a}{2}+5a=\frac{13a}{2}

So the other intersection is

\left(\frac{3a}{2},\frac{13a}{2}\right)

Therefore the two points are

\boxed{\left(-\frac{5a}{4},\frac{15a}{4}\right) \quad \text{and}\quad \left(\frac{3a}{2},\frac{13a}{2}\right)}

IAL 2021 Jan Q4

题目

解答

(a)

(b)

(c)

Left intersection

Right intersection