Question:

Given f(x)=x^2+3x

x such that f(x)=4 (ie y=4, what is x?)

I am honestly very confused by this question to even attempt to do it.

Though here is what I have on the rest of my sheet as my answers. Please correct me if I'm wrong. Thank you!

Given f(x)=x^2+3x

A. f(2)= 10
B. 3f(-1)=6
C. f(0)+f(2)= 10
D. f(t+1)= t^2+5t+4
E. x such that f(x)=4 (ie y=4 what is x?)
Don't have an answer for this one...

A, C, and D are correct

B:
3f(-1) = 3[(-1)^2 + 3(-1)]
= 3[1 - 3]
= 3(-2) = -6

E)
x^2 + 3x = 4
x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4 or x = 1

checking for x = -4
f(-4) = (-4)^2 + 3(-4)
= 16-12 = 4 , checks!

try for x = 1 the same way, you should get 4 as well

Oh my god, thank you so so much! :-)

To find the value of x such that f(x) = 4, you need to solve the equation f(x) = 4.

Given that f(x) = x^2 + 3x, you can substitute this expression into the equation:

x^2 + 3x = 4

Now, you need to solve this quadratic equation for x. Rearrange the equation to have zero on one side:

x^2 + 3x - 4 = 0

To solve this quadratic equation, you can either factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 1, b = 3, and c = -4.

x = (-3 ± √(3^2 - 4 * 1 * -4)) / (2 * 1)

Simplifying further:

x = (-3 ± √(9 + 16)) / 2

x = (-3 ± √25) / 2

x = (-3 ± 5) / 2

Now, we have two possible solutions:

x = (-3 + 5) / 2 = 2 / 2 = 1

x = (-3 - 5) / 2 = -8 / 2 = -4

Therefore, the values of x such that f(x) = 4 are x = 1 and x = -4.

Regarding the other options:

A. f(2) = 2^2 + 3(2) = 4 + 6 = 10 (correct)
B. 3f(-1) = 3(-1)^2 + 3(-1) = 3 + (-3) = 0 (incorrect, it should be 0, not 6)
C. f(0) + f(2) = 0^2 + 3(0) + 2^2 + 3(2) = 0 + 0 + 4 + 6 = 10 (correct)
D. f(t+1) = (t+1)^2 + 3(t+1) = t^2 + 2t + 1 + 3t + 3 = t^2 + 5t + 4 (correct)
E. x such that f(x) = 4, as discussed above, the solutions are x = 1 and x = -4.