Question

Two functions are shown in the table below. Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = −x + 6 Complete the table on your own paper, then select the value that is a solution to f(x) = g(x).

Answer 1

For $\fbox{\begin \\\math{x}=6\\\end{minispace}}$ the function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ has same value.

Step by step explanation:  

The given functions are,

$f(x)=-x^{2}+4x+12$

$g(x)=-x+6$

Step 1:  

Substitute $x=1$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(1)$.

$f(1)=-1^{2} +4(1)+12\\f(1)=-1+4+12\\f(1)=15$

Substitute $x=1$ in $g(x)=-x+6$ to obtain the value of $g(1)$ .

$g(1)=-1+6\\g(1)=5$

Step 2:

Substitute $x=2$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(2)$.

$f(2)=-2^{2} +4(2)+12\\f(2)=-4+8+12\\f(2)=16$

Substitute $x=2$ in $g(x)=-x+6$ to obtain the value of $g(2)$ .

$g(2)=-2+6\\g(2)=4$

Step 3:

Substitute $x=3$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(3)$.

$f(3)=-3^{2} +4(3)+12\\f(3)=-9+12+12\\f(3)=15$

Substitute $x=3$ in $g(x)=-x+6$ to obtain the value of $g(3)$ .

$g(3)=-3+6\\g(3)=3$

Step 4:

Substitute $x=4$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(4)$.

$f(4)=-4^{2} +4(4)+12\\f(4)=-16+16+12\\f(4)=12$

Substitute $x=4$ in $g(x)=-x+6$ to obtain the value of $g(4)$ .

$g(4)=-4+6\\g(4)=2$

Step 5:

Substitute $x=5$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(5)$.

$f(5)=-5^{2} +4(5)+12\\f(5)=-25+20+12\\f(5)=7$

Substitute $x=5$ in $g(x)=-x+6$ to obtain the value of $g(5)$ .

$g(5)=-5+6\\g(5)=1$

Step 6:

Substitute $x=6$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(6)$.

$f(6)=-6^{2} +4(6)+12\\f(6)=-36+24+12\\f(6)=0$

Substitute $x=6$ in $g(x)=-x+6$ to obtain the value of $g(6)$ .

$g(6)=-6+6\\g(6)=0$

Step 7:

As per the given condition $f(x)=g(x)$.

(a). Substitute $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ in above equation.

$-x^{2} +4x+12=-x+6$

(b). Multiply with $-1$ on both sides.

$x^{2} -4x-12=x-6$

(c). Shift the term $x-6$ to left hand side.

$x^{2} -4x-12-x+6=0\\x^{2} -5x-6=0$

(d). Split the middle term in such a way that its sum is 5 and multiplication is 6.

$x^{2} -(6-1)x-6=0\\x^{2} -6x+x-6=0\\x(x-6)+1(x-6)=0\\(x+1)(x-6)=0\\x=-1 ,6$

It is observed from the above solution that for $x=6$ both the functions $f(x)$ and $g(x)$ has same value.

Direct method:

$f(x)=g(x)\\\Leftrightarrow-x^{2} +4x+12=-x+6\\\Leftrightarrow-x^{2} +4x+12+x-6=0\\\Leftrightarrow-x^{2} +5x+6=0\\\Leftrightarrow-x^{2} +6x-x+6=0\\\Leftrightarrow x^{2} -6x+x-6=0\\\Leftrightarrow x(x-6)+1(x-6)=0\\\Leftrightarrow(x+1)(x-6)=0\\\Leftrightarrow x=6,-1$

The table for the function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ is attached below.

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Answer details:  

Grade: Middle school.  

Subjects: Mathematics.  

Chapter: Function.

Keywords: Function, Middle term split method, Binomial,Quadratic, Polynomial, Factorized, Perfect square, Zeros, Zeros of a function, Expression, Equation, x, x^2, x^3, -x^2+4x+12, -x+6, roots of equation.

Answer 2

Answer:

The answer would be D.

Step-by-step explanation: