SBMPTN Zone : Fungsi Kuadrat [2]

Berikut ini adalah kumpulan soal mengenai Fungsi Kuadrat tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tipe:

Standar SBMPTN HOTS

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No. 11

Parabola {y=ax^2+bx+c}, puncaknya {(p,q)} dicerminkan terhadap garis {y=-1} menghasilkan {y=x^2+x-2}. Nilai {a+b+c+p+q} adalah

1. -\dfrac92
2. -\dfrac94
3. -2

4. \dfrac94
5. \dfrac92

Bisa dikatakan bahwa {y=ax^2+bx+c} adalah bayangan {y=x^2+x-2} setelah dicerminkan terhadap garis {y=-1}.
x'=x\rightarrow x=x'
y'=2(-1)-y=-2-y\rightarrow y=-2-y'

$\begin{array}{rl}y& =x^2+x-2\\ -2-y^{\prime }& =(x^{\prime }{)}^2+x^{\prime }-2\\ -y^{\prime }& =(x^{\prime }{)}^2+x^{\prime }\\ y^{\prime }& =-(x^{\prime }{)}^2-x^{\prime }\\ y& =-x^2-x\end{array}$
a=-1, b=-1, c=0

$\begin{array}{rl}p& =-{\displaystyle \frac{b}{2a}}\\ & =-{\displaystyle \frac{-1}{2(-1)}}\\ & ={\displaystyle \frac{1}{2}}\end{array}$

$\begin{array}{rl}q& =-p^2-p\\ & =-{\left({\displaystyle \frac{1}{2}}\right)}^2-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2}}\\ & =-{\displaystyle \frac{3}{4}}\end{array}$

$\begin{array}{rl}a+b+c+p+q& =-1+(-1)+0+{\displaystyle \frac{1}{2}}+(-{\displaystyle \frac{3}{4}})\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{9}{4}}}}}}\end{array}$

Jadi, {a+b+c+p+q=-\dfrac94}.
JAWAB: B

No. 12

Agar grafik fungsi {y = x^2 + 2x-5} seluruhnya di bawah grafik fungsi {y = 2x^2-ax + 2a + 4} maka nilai a

1. -4\lt a\lt8
2. -8\lt a\lt4
3. a\lt-8\text{ atau }a\gt-4

4. a\lt8
5. a\gt-4

SKMP Juli 2020

$\begin{array}{rl}{x}^2+2x-5& <2x^2-ax+2a+4\\ -x^2+(a+2)x-9& <0\end{array}$
{a=-1}, {b=a+2}, {c=-9}

$\begin{array}{rl}D& <0\\ b^2-4ac& <0\\ (a+2{)}^2-4(-1)(-9)& <0\\ a^2+4a+4-36& <0\\ a^2+4a-32& <0\\ (a+8)(a-4)& <0\end{array}$

-8\lt a\lt4

Jadi, -8\lt a\lt4.
JAWAB: B

No. 13

Diketahui f(x) dan g(x) memenuhi:
{f(x) + 3g(x) = x^2-x-2}
{2f(x) + 4g(x) = 2x^2-8}
Untuk semua nilai x, jika x_1 dan x_2 memenuhi {f(x) = g(x)} maka nilai x_1\cdot x_2 adalah

1. -10
2. -11
3. 12

4. 11
5. 10

SKMP Agustus 2020

2f(x) + 4g(x) = 2x^2-8\qquad{\color{red}{:2}}

$\begin{array}{rl}f(x)+3g(x)& =x^2-x-2\\ f(x)+2g(x)& =x^2-4& -\\ g(x)& =-x+2\end{array}$

$\begin{array}{rl}f(x)+2g(x)& =x^2-4\\ f(x)+2(-x+2)& =x^2-4\\ f(x)-2x+4& =x^2-4\\ f(x)& =x^2+2x-8\end{array}$

$\begin{array}{rl}f(x)& =g(x)\\ x^2+2x-8& =-x+2\\ x^2+3x-10& =0\end{array}$

$\begin{array}{rl}{x}_1\cdot x_2& ={\displaystyle \frac{c}{a}}\\ & ={\displaystyle \frac{-10}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -10}}}}\end{array}$

Jadi, x_1\cdot x_2=-10. JAWAB: A

No. 14

Parabola y=x^2-6x+18 digeser ke kanan sejauh a dan digeser ke bawah sejauh 6a satuan. Jika hasil pergeseran parabola ini memotong sumbu x di satu titik. maka nilai 6a+5=

1. 3
2. 4
3. 8

4. 11
5. 14

Telegram

$\begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}a\\ -6a\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& =\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)-\left(\begin{array}{c}a\\ -6a\end{array}\right)\\ & =\left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }+6a\end{array}\right)\end{array}$
x=x'-a
y=y'+6a

$\begin{array}{rl}y& =x^2-6x+18\\ y^{\prime }+6a& ={(x^{\prime }-a)}^2-6(x^{\prime }-a)+18\\ y^{\prime }+6a& ={\left(x^{\prime }\right)}^2-2a{x}^{\prime }+a^2-6x^{\prime }+6a+18\\ y^{\prime }& ={\left(x^{\prime }\right)}^2+(-2a-6)x^{\prime }+a^2+18\\ y& =x^2+(-2a-6)x+a^2+18\end{array}$

memotong di satu titik → D=0
$\begin{array}{rl}D& =0\\ b^2-4ac& =0\\ (-2a-6{)}^2-4(1)(a^2+18)& =0\\ 4a^2+24a+36-4a^2-72& =0\\ -24a-36& =0\\ -24a& =36\\ 6a& =-{\displaystyle \frac{36}{4}}\\ & =9\\ 6a+5& =9+5\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 14}}}}\end{array}$

Jadi, 6a+5=14.
JAWAB: E

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