Question

How to do Question 3? Do I square y?

Answer 1

Explanation:

a. To write the third equation using the point (2, 19), we substitute the values into the equation: 19 = 4a + 2b + c

b. To solve the system of three linear equations, we can use various methods such as substitution or elimination. Let's use the elimination method:

First, we can subtract the first equation from the second equation to eliminate c: (10 - 5) = (4a + 2b + c) - (a + b + c) 5 = 3a + b

Next, we can subtract the first equation from the third equation to eliminate c: (19 - 5) = (4a + 2b + c) - (a + b + c) 14 = 3a + b

Now we have two equations: 5 = 3a + b 14 = 3a + b

Subtracting the first equation from the second equation, we get: (14 - 5) = (3a + b) - (3a + b) 9 = 0

This results in a contradiction, indicating that there is no solution to the system of equations. Geometrically, this means that the three points (1, 5), (2, 10), and (2, 19) do not lie on the same curve.

If we correct the third point to (3, 19), the equation becomes: 19 = 9a + 3b + c

To solve the new system of equations, we can use the same elimination method:

Subtracting the first equation from the second equation, we get: (10 - 5) = (4a + 2b + c) - (a + b + c) 5 = 3a + b

Subtracting the first equation from the third equation, we get: (19 - 5) = (9a + 3b + c) - (a + b + c) 14 = 8a + 2b

Now we have two equations: 5 = 3a + b 14 = 8a + 2b

Multiplying the first equation by 2 and subtracting it from the second equation, we get: (14 - 2(5)) = (8a + 2b) - (6a + 2b) 4 = 2a

Dividing both sides by 2, we find: a = 2

Substituting the value of a into the first equation, we can solve for b: 5 = 3(2) + b 5 = 6 + b b = -1

Substituting the values of a and b into the equation 5 = a + b + c, we can solve for c: 5 = 2 + (-1) + c 5 = 1 + c c = 4

So the quadratic function is: y = 2x^2 - x + 4

To find all values for t that will change the quadratic function y = ax^2 + bx + c into a linear function, we set the coefficient of x^2 to zero:

0 = a a = 0

This means that any value of a equal to zero will result in a linear function.

Answer 2

`\textsf{1.a.} \quad 19=4a+2b+c`

`\textsf{1.b.} \quad \sf No\;solutions`

`\textsf{2.} \quad\;\; 19=9a+3b+c, \;\;\;y=2x^2-x+4`

`\textsf{3.} \quad\;\; t=15`

Explanation:

Question 1

To use the point (2, 19) to write the third equation, substitute x = 2 and y = 19 into y = ax² + bx + c:

`19=a(2)^2+b(2)+c`

`19=4a+2b+c`

Therefore, the system of equations is:

`\begin{cases}5=a+b+c\\10=4a+2b+c\\19=4a+2b+c\end{cases}`

The system cannot be solved as the third equation contradicts the second equation. Both equations have the same left-hand side (4a + 2b + c), but different right-hand sides (10 and 19).

`\hrulefill`

Question 2

To rewrite the third equation with the correct third point (3, 19), substitute substitute x = 3 and y = 19 into y = ax² + bx + c:

`19=a(3)^2+b(3)+c`

`19=9a+3b+c`

Therefore, the correct system of equations is:

`\begin{cases}5=a+b+c\\10=4a+2b+c\\19=9a+3b+c\end{cases}`

To solve the system of equations, subtract the second equation from the third equation to eliminate c:

`\begin{array}{crcccccl}&9a&+&3b&+&c&=&19\\-&(4a&+&2b&+&c&=&10)\\\cline{2-8}&5a&+&b&&&=&\;\;9\end{array}`

Rearrange to isolate b:

`b=9-5a`

Subtract the first equation from the second equation to eliminate c:

`\begin{array}{crcccccl}&4a&+&2b&+&c&=&10\\-&(a&+&b&+&c&=&5)\\\cline{2-8}&3a&+&b&&&=&5\end{array}`

Rearrange to isolate b:

`b=5-3a`

Substitute one of the equations for b into the other and solve for a:

`\begin{aligned}5-3a&=9-5a\\2a&=4\\a&=2\end{aligned}`

Substitute the found value of a into one of the equations for b and solve for b:

`\begin{aligned}b&=5-3(2)\\b&=5-6\\b&=-1\end{aligned}`

Substitute the found values of a and b into the first equation and solve for c:

`\begin{aligned}a+b+c&=5\\2-1+c&=5\\1+c&=5\\c&=4\end{aligned}`

Therefore, the values of a, b and c are:

• a = 2
• b = -1
• c = 4

Therefore, the quadratic equation is:

`y=2x^2-x+4`

`\hrulefill`

Question 3

If the function is a linear function then the three points will lie on a straight line.

Use the first two points (1, 5) and (2, 10) to find the slope of the line by substituting them into the slope formula:

`\textsf{Slope}=(y_2-y_1)/(x_2-x_1)=(10-5)/(2-1)=(5)/(1)=5`

Substitute the found slope and one of the points (1, 5) into the slope-intercept form of a linear equation:

`\begin{aligned}y-y_1&=m(x-x_1)\\y-5&=5(x-1)\\y-5&=5x-5\\y&=5x\end{aligned}`

Therefore, the equation of the linear function is:

`y=5x`

To find the value of t that changes the quadratic into a linear function, substitute (3, t) into the linear function:

`t=5(3)`

`t=15`

Therefore, the value of t is 15.