Given Vertex of Quadratic Equation Find the Sum of b and c

[etoday]

how do you solve if there was no value for a
something more like
ax²+bx+c
with the same vertex

[herbcruz]

Here's another way to solve this problem.

Standard Form: y = ax^2+bx+c, where a is the coefficient of the x^2-term, b is the coefficient of the x-term, and c is the constant term.

Vertex Form: y = a*(x-h)^2 + k, where the point, (h, k), is the vertex of the parabola

Our Equation: y = 3x^2+bx+c, which is in Standard Form, and from this, we know that a = 3, and we also know that the vertex of the parabola is the point, (-1, 4) = (h, k) (i.e., h = -1 and k = 4), so we can re-write this equation in Vertex Form.

y = 3*(x-(-1))^2+4 = 3*(x+1)^2+4.

Hence, y = 3*(x+1)^2+4.

We'll re-expand this, now.

Knowing how to expand a Binomial Sum when it's squared, we know that (a+b)^2 = a^2+2ab+b^2.

By this, we get that (x+1)^2 = (x)^2+2*(x)*(1)+(1)^2 = x^2+2x+1

(Scratch Work: (x)^2 = x^2; 2*(x)*(1) = 2x; (1)^2 = 1 (because (1)^2 = 1^2 = 1*1 = 1))

So, (x+1)^2 = x^2+2x+1.

y = 3*(x+1)^2+4

Expanding (x+1)^2 (The work that shows this expansion is above), we get the following:

y = 3*(x^2+2x+1)+4

Distribute the 3.

y = 3x^2+6x+3+4

Combine like terms.

y = 3x^2+6x+7.

And now, we have our quadratic function in standard form, again.

Comparing y = ax^2+bx+c with y = 3x^2+6x+7, we get that the values for b and c are 6 and 7, respectively (i.e., b = 6 and c = 7).

The problem asks us to find the sum of b and c (i.e., the value of b+c).

Therefore, b+c = 6+7 = 13.

So, b+c = 13.

The final answer is 13, and we are now finished with this problem.

The correct answer is choice C..