题目

Problem

(a) Express $3x^2+12x+13$ in the form

\begin{align*} a(x+b)^2+c \end{align*}

where $a$ , $b$ and $c$ are integers to be found.

(3)

(b) Hence sketch the curve with equation $y=3x^2+12x+13$ .

On your sketch show clearly

(3)

解答

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先提出 $3$ ，再完成平方。

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\begin{align*} 3x^2+12x+13 =&\,3(x^2+4x)+13\\ =&\,3\left((x+2)^2-4\right)+13\\ =&\,3(x+2)^2-12+13\\ =&\,3(x+2)^2+1. \end{align*}

Therefore

\begin{align*} a=3,\qquad b=2,\qquad c=1. \end{align*}

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由 $3(x+2)^2+1$ 可知顶点是 $(-2,1)$ ，且开口向上。 $y$ 截距由 $x=0$ 得到。

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The curve is

\begin{align*} y=3(x+2)^2+1. \end{align*}

So the turning point is

\begin{align*} (-2,1). \end{align*}

Since the coefficient of $(x+2)^2$ is positive, the curve opens upwards.

The $y$ -intercept occurs when $x=0$ :

\begin{align*} y=3(0)^2+12(0)+13=13. \end{align*}

So the $y$ -intercept is

\begin{align*} (0,13). \end{align*}

The sketch should show an upward-opening parabola with turning point $(-2,1)$ and $y$ -intercept $(0,13)$ .