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ziro4ka
12 days ago
6

Grace drew a triangle. It's sides were 12 mm, 10 mm, and 11 mm. It has one obtuse angle and two acute angles. What triangle did

she draw?
Mathematics
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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 o
Svet_ta [12734]

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

Detailed breakdown:  

The functions involved are

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

g(x)=-x+6

Step 1:  

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

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

Insert x=1 in g(x)=-x+6 to find the value of g(1).

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

Step 2:

Insert 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 into g(x)=-x+6 to find the value of g(2).

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

Step 3:

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

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

Also, replace x=3 in g(x)=-x+6 to find the value of g(3).

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

Step 4:

Insert x=4 in f(x)=-x^{2} +4x+12 to find the value of f(4).

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

Also, replace x=4 in g(x)=-x+6 to obtain the value of g(4).

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

Step 5:

Insert 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

Replace x=5 in g(x)=-x+6 to find the value of g(5).

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

Step 6:

Insert x=6 into f(x)=-x^{2} +4x+12 to find the value of f(6).

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

Also, substitute x=6 in g(x)=-x+6 to obtain the value of g(6).

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

Step 7:

According to the provided condition f(x)=g(x).

(a). Insert f(x)=-x^{2} +4x+12 and g(x)=-x+6 into the previously mentioned equation.

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

(b). Multiply through by -1 on both sides.

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

(c). Move the term x-6 to the left side of the equation.

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

(d). Divide the middle term so that its sum equals 5 and the product equals 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

From the analysis above, it is noted that for x=6 both functions f(x) and g(x) yield the same outcome.

Using a direct approach:

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 representing function f(x)=-x^{2} +4x+12 and g(x)=-x+6 is included below.

For more information:

1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)

2. Which is the graph of f(x) = (x – 1)(x + 4)?

6 0
1 month ago
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