Write the equation of the parabola that has a vertex (1, -1) and passes through the point (2,5)

Do you know the "vertex form" of a quadratic equation?

no

Vertex Form: \( y - k = a(x - h) \) Although some people learn it as standard form. I learned it as "Vertex form" though. If we use that form, the vertex can be put in as (h,k) = (1,-1) Then, we just have to find a by using the given point as the values for (x,y), and solving for a.

what did u get

y - (-1) = a(x - 1) <== (x,y) = (2,5) y + 1 = a(x - 1) 5 + 1 = a(2 - 1) 6 = a y + 1 = 6(x - 1) y = 6(x-1) - 1 But, if you don't know vertex form, I think we could also use \(ax^2 + bx + c = y\) and some relations of the coefficients...

Yeah. y = ax^2 + bx + c We can create a system of equations using the two points and a relation between the vertex and the coefficients. 5 = a(2)^2 + b(2) + c ==> 5 = 4a + 2b + c } System -1 = a(1)^2 + b(1) + c ==> 1 = a + b + c } -b/2a = 1 (x-value of vertex is -b/2a) -b = 2a b = -2a Substitute into the system to eliminate (b) 5 = 4a + 2(-2a) + c -1 = a + (-2a) + c 5 = 4a - 4a + c -1 = a - 2a + c 5 = c --, Substitute C -1 = -a + c -1 = -a + 5 -6 = -a a = 6 ==> b = -2(6) = -12 y = 6x^2 -12x + 5

Which, if we compare with the vertex-form equation, is the same if we expand the binomial and distribute/ simplify: y = 6(x - 1)^2 - 1 = 6(x^2 - 2x + 1) - 1 = 6x^2 - 12x + 6 - 1 = 6x^2 - 12x + 5

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