Question 1201024
the vertex form of the equation is y = a * (x-h)^2 + k.
(h,k) is the vertex.
when the vertex is (1,2), this becomes y = a * (x-1)^2 + 2
when x = 0, y = 0.
therefore you get:
0 = a * (-1)^2 + 2
simplify to get:
0 = a + 2
solve for a to get:
a = -2
vertex form of the equation becomes y = -2 * (x-1)^2 + 2.
looks like the coefficient of the x^2 term is -2
the standard form of the equation is y = ax^2 + bx + c
to convert to this form, set y = 0 in the vertex form and solve.
you get 0 = -2 * (x-1)^2 + 2
simplify to get:
0 = -2 * (x^2 -2x + 1) + 2
simpify to get:
0 = -2*x^2 + 4x -2 + 2
simplify to get:
0 = -2x^2 + 4x.
that's the standard form of the equation.
your solution is that the coefficient of the x^2 term is -2.
here's the graph.


<img src = "http://theo.x10hosting.com/2023/031604.jpg">


both forms of the equation show on the graph.
they both show the same figure on the graph, indicating that they are equivalent to each other.


here's a reference.


<a href = "https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html" target = "_blank">https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html</a>