# Tutor profile: Brad P.

## Questions

### Subject: Linear Algebra

Let v1 = [1 3 2], v2 = [-1 2 -1], and v3= [5 5 8]. Determine whether the given vectors are linearly independent.

If a set of vectors is linearly independent, then there is only the trivial solution to the homogeneous system. That is, if a(v1) + b(v2) + c(v3) = [0 0 0] only if a=b=c=0, then v1, v2, and v3 are linearly independent. In order to solve this system, we create the following 3x3 matrix: ( 1 -1 5 ) ( 3 2 5 ) ( 2 -1 8 ) There is no need to augment the matrix with the zero vector as it will not be affected by row operations. Next, we row reduce to find the solution to the system. Row reduction yields the following matrix in Row-Echelon form: ( 1 -1 5 ) ( 0 1 -2 ) ( 0 0 0 ) Since there is a zero row, have a parameter which can be any real number. This means there are infinitely many non-trivial solutions to the homogeneous system, meaning the vectors are linearly dependent.

### Subject: Calculus

Let f(x) = ax^3 + bx^2 where a and b are constants. Find a and b such that f(x) has a point of inflection at (1,3).

Firstly, we see that f(1)=3. So, we can plug in to find the equation 3 = a(1)^3 + b(1)^2, or a + b = 3. Next, we take the second derivative of the function, and set it equal to 0. This allows us to relate the inflection point to a and b. We find that f'(x) = 3ax^2 + 2bx, and then f''(x) = 6ax + 2b by using the power rule. Since the point of inflection is at x=1, f''(1)=0. So, we are given the equation 6a(1) + 2b = 0, or 6a + 2b = 0. Now, we have 2 equations and two unknowns, so we can solve using substitution or elimination. Using substitution, we can find that 6a + 2b = 0 implies that a = -(1/3)b. So, -(1/3)b + b =3, by plugging in to the other equation. This gives us b=9/2. By using b=9/2, a + 9/2 = 3, which implies a = -3/2. So, if a=-3/2 and b= 9/2, then f(x) will have an inflection point at (1,3).

### Subject: Algebra

A man throws a ball such that it's height (y) at a given x-position is given by the function y = -x^2 - 5x. If the man was standing at the origin when he threw the ball, at what x-value does the ball hit the ground?

We see that if we factor the given function, we get y = -x(x-5). By, setting y=0, we can find when the ball will be on the ground (the roots of the function). So, 0= -x(x-5). We note that if x=0 or if (x-5)=0, then the product -x(x-5)=0. It follows that the ball is on the ground at x=0 and x=5. We are given the man throws the ball from x=0 (the origin), so the ball must land at x=5.

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