Let g(x)=x2+2(a+b)x+2(a2+b2),where a and b are constants. b. Use the result in part (a) to explain why the graph of g has no x -intercepts unless a=ba = b, in which case there is only one x-intercept.

Pertanyaan

Let $g\left(x\right)=x^2+2\left(a+b\right)x+2\left(a^2+b^2\right)$ ,where a and b are constants.

b. Use the result in part (a) to explain why the graph of $g$ has no x -intercepts unless $a=b$ a = b, in which case there is only one $x$ -intercept.

I. Sutiawan

Master Teacher

Mahasiswa/Alumni Universitas Pasundan

Jawaban terverifikasi

Jawaban

dapat disimpulkan ketika nilai a equals b

, maka grafik g open parentheses x close parentheses equals x squared plus 2 open parentheses a plus b close parentheses x plus 2 open parentheses a squared plus b squared close parentheses

g open parentheses x close parentheses equals x squared plus 2 open parentheses a plus b close parentheses x plus 2 open parentheses a squared plus b squared close parentheses

melalui sumbu

di satu titik (menyinggung sumbu

Pembahasan

Diketahui g open parentheses x close parentheses equals x squared plus 2 open parentheses a plus b close parentheses x plus 2 open parentheses a squared plus b squared close parentheses , maka untuk menentukan vertex atau titik puncaknya kita dapat menggunakan rumus $left parenthesis x subscript p comma space y subscript p right parenthesis equals open parentheses negative fraction numerator b over denominator 2 a end fraction comma space minus fraction numerator D over denominator 4 a end fraction close parentheses$ , sehingga:

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript p end cell equals cell negative fraction numerator b over denominator 2 a end fraction end cell row blank equals cell negative fraction numerator 2 left parenthesis a plus b right parenthesis over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row blank equals cell negative fraction numerator 2 left parenthesis a plus b right parenthesis over denominator 2 end fraction end cell row blank equals cell negative left parenthesis a plus b right parenthesis end cell row blank blank blank row cell y subscript p end cell equals cell negative fraction numerator D over denominator 4 a end fraction end cell row blank equals cell negative fraction numerator b squared minus 4 a c over denominator 4 a end fraction end cell row blank equals cell negative fraction numerator left parenthesis 2 left parenthesis a plus b right parenthesis right parenthesis squared minus 4 left parenthesis 1 right parenthesis left parenthesis 2 left parenthesis a squared plus b squared right parenthesis right parenthesis over denominator 4 left parenthesis 1 right parenthesis end fraction end cell row blank equals cell negative fraction numerator 4 left parenthesis a squared plus 2 a b plus b squared right parenthesis minus 8 a squared minus 8 b squared over denominator 4 end fraction end cell row blank equals cell negative fraction numerator 4 a squared plus 8 a b plus 4 b squared minus 8 a squared minus 8 b squared over denominator 4 end fraction end cell row blank equals cell negative fraction numerator negative 4 a squared plus 8 a b minus 4 b squared over denominator 4 end fraction end cell row blank equals cell negative fraction numerator up diagonal strike 4 left parenthesis negative a squared plus 2 a b minus b squared right parenthesis over denominator up diagonal strike 4 end fraction end cell row blank equals cell a squared minus 2 a b plus b squared end cell row blank equals cell left parenthesis a minus b right parenthesis squared end cell end table$

Dari perhitungan di atas, titik puncak dari fungsi kuadrat g open parentheses x close parentheses equals x squared plus 2 open parentheses a plus b close parentheses x plus 2 open parentheses a squared plus b squared close parentheses adalah left parenthesis negative left parenthesis a plus b right parenthesis comma space left parenthesis a minus b right parenthesis squared right parenthesis .

Ketika , maka titik puncak menjadi:

Karena ordinat titik puncak bernilai 0, artinya grafik menyinggung sumbu ketika titik puncak. Dalam parabola, jika grafik memotong sumbu di titik puncak, maka grafik tersebut hanya menyinggung dan tidak mungkin memotong di dua titik.

Sehingga dapat disimpulkan ketika nilai , maka grafik melalui sumbu di satu titik (menyinggung sumbu ).