1. ## Linear Functions

• Given P(x) = x^2+7x+c and P(-3)=3, find c.

If I were to do this i would sub in -3 for x so...
3=-3^2+(7x3)+c
3 = 9+21+c
3 = 30+c
c=-27
But im not sure if this is right?

Also
• for the linear function f. It is known that f(1)=1 and f(-2)=-8. Find the function rule.
I am a little stuck with this one and unsure how to approach it??

2. Originally Posted by scubasteve94
• Given P(x) = x^2+7x+c and P(-3)=3, find c.

If I were to do this i would sub in -3 for x so...
3=-3^2+(7x3)+c
3 = 9+21+c
3 = 30+c
c=-27
But im not sure if this is right?
almost. you will have 3 = 9 - 21 + C

Also
• for the linear function f. It is known that f(1)=1 and f(-2)=-8. Find the function rule.

I am a little stuck with this one and unsure how to approach it??
note that you have the points (1,1) and (-2,-8) on the line. you can use these to find the slope of the line (do you know how?), once you have that, you can find the equation of the line, which is the rule you are looking for. it will be of the form y = mx + b, where m is the slope, and b is the y-intercept

3. Originally Posted by Jhevon
almost. you will have 3 = 9 - 21 + C
so it will be
3 = 9 -21 + C
3 = -12 + C
C = 15

Originally Posted by Jhevon
note that you have the points (1,1) and (-2,-8) on the line. you can use these to find the slope of the line (do you know how?), once you have that, you can find the equation of the line, which is the rule you are looking for. it will be of the form y = mx + b, where m is the slope, and b is the y-intercept
so
y = mx + b
y = (1/3)x + b
sub in point (1,1) to find b
1 = (1/3)x1+b
b = (2/3)
therefore f(x) = (1/3)x + (2/3)

4. Originally Posted by scubasteve94
so it will be
3 = 9 -21 + C
3 = -12 + C
C = 15
yes

so
y = mx + b
y = (1/3)x + b
sub in point (1,1) to find b
1 = (1/3)x1+b
b = (2/3)
therefore f(x) = (1/3)x + (2/3)
no. recall the slope formula for the slope between points $(x_1,y_1)$ and $(x_2,y_2)$. it is: $\frac {y_2 - y_1}{x_2 - x_1}$

5. Originally Posted by Jhevon
no. recall the slope formula for the slope between points $(x_1,y_1)$ and $(x_2,y_2)$. it is: $\frac {y_2 - y_1}{x_2 - x_1}$
so it would be
y = mx + b
y = 3x + b
sub point (1,1) to find b
1 = 3x1 + b
1 = 3 + b
b = -2
therefore f(x) = 3x - 2

6. Originally Posted by scubasteve94
so it would be
y = mx + b
y = 3x + b
sub point (1,1) to find b
1 = 3x1 + b
1 = 3 + b
b = -2
therefore f(x) = 3x - 2
yes

7. thank you so much
you were a lot of help