I learned a bit of calculus a good while ago but am honestly a bit embarrassed not to know the answer to this question, but a friend said it was much easier to solve a quadratic equation with calculus than to use the quadratic formula, I just wanted to understand how
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A quadratic function describes a parabola.
Calculus is an easy method for finding the x-value of the vertex.
The quadratic formula is really not useful for finding the vertex. The algebraic method for finding the vertex is "completing the square."
The quadratic formula (or factoring, or completing the square) is a way to find the "zeros" of the function.
Calculus is not particularly useful for finding zeros.
the way calculus is used to find the vertex is based on the fact that the slope of the vertex is zero, so by taking the first derivative of the quadratic function, it is very simple to find that point.
the algebraic method for finding the vertex is to manipulate
y = ax² + bx + c
into
y = a(x - h)² + k
where (h, k) is the vertex
♣♦
Calculus is an easy method for finding the x-value of the vertex.
The quadratic formula is really not useful for finding the vertex. The algebraic method for finding the vertex is "completing the square."
The quadratic formula (or factoring, or completing the square) is a way to find the "zeros" of the function.
Calculus is not particularly useful for finding zeros.
the way calculus is used to find the vertex is based on the fact that the slope of the vertex is zero, so by taking the first derivative of the quadratic function, it is very simple to find that point.
the algebraic method for finding the vertex is to manipulate
y = ax² + bx + c
into
y = a(x - h)² + k
where (h, k) is the vertex
♣♦