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Accedmychevron_right11thchevron_rightmathchevron_rightTheory Of Quadratic Functionschevron_rightExercise 3.1

Solution

We complete the square to write each quadratic in vertex form:

f(x)=a(x−h)2+k f(x)=a(x-h)^2+k

Then:

  • If a>0a>0, the minimum value is kk.
  • If a<0a<0, the maximum value is kk.

(i) f(x)=x2+6x+13f(x)=x^2+6x+13

f(x)=x2+6x+13=(x2+6x+9)+4=(x+3)2+4\begin{aligned} f(x)&=x^2+6x+13\\ &=\left(x^2+6x+9\right)+4\\ &=(x+3)^2+4 \end{aligned} min⁡f(x)=4 at x=−3\boxed{\min f(x)=4\text{ at }x=-3}

(ii) f(x)=x2+4xf(x)=x^2+4x

f(x)=x2+4x=(x2+4x+4)−4=(x+2)2−4\begin{aligned} f(x)&=x^2+4x\\ &=\left(x^2+4x+4\right)-4\\ &=(x+2)^2-4 \end{aligned} min⁡f(x)=−4 at x=−2\boxed{\min f(x)=-4\text{ at }x=-2}

(iii) f(x)=−x2+8x+13f(x)=-x^2+8x+13

f(x)=−(x2−8x)+13=−(x2−8x+16)+16+13=−(x−4)2+29\begin{aligned} f(x)&=-\left(x^2-8x\right)+13\\ &=-\left(x^2-8x+16\right)+16+13\\ &=-(x-4)^2+29 \end{aligned} max⁡f(x)=29 at x=4\boxed{\max f(x)=29\text{ at }x=4}

(iv) f(x)=−x2−3x−5f(x)=-x^2-3x-5

f(x)=−(x2+3x+5)=−[(x+32)2+114]=−(x+32)2−114\begin{aligned} f(x)&=-\left(x^2+3x+5\right)\\ &=-\left[\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\right]\\ &=-\left(x+\frac{3}{2}\right)^2-\frac{11}{4} \end{aligned} max⁡f(x)=−114 at x=−32\boxed{\max f(x)=-\frac{11}{4}\text{ at }x=-\frac{3}{2}}

(v) f(x)=3x2+6x−13f(x)=3x^2+6x-13

f(x)=3(x2+2x)−13=3[(x+1)2−1]−13=3(x+1)2−16\begin{aligned} f(x)&=3(x^2+2x)-13\\ &=3\left[(x+1)^2-1\right]-13\\ &=3(x+1)^2-16 \end{aligned} min⁡f(x)=−16 at x=−1\boxed{\min f(x)=-16\text{ at }x=-1}

(vi) f(x)=−2x2−x+21f(x)=-2x^2-x+21

f(x)=−2(x2+12x)+21=−2[(x+14)2−116]+21=−2(x+14)2+18+21=−2(x+14)2+1698\begin{aligned} f(x)&=-2\left(x^2+\frac{1}{2}x\right)+21\\ &=-2\left[\left(x+\frac{1}{4}\right)^2-\frac{1}{16}\right]+21\\ &=-2\left(x+\frac{1}{4}\right)^2+\frac{1}{8}+21\\ &=-2\left(x+\frac{1}{4}\right)^2+\frac{169}{8} \end{aligned} max⁡f(x)=1698 at x=−14\boxed{\max f(x)=\frac{169}{8}\text{ at }x=-\frac{1}{4}}
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