Question

Given the vertex ( 4, 6 )and the point(1, -3), write the vertex form equation of the quadratic

Answer 1

Answer:

$\displaystyle f(x) = -(x-4)^2 + 6$

Step-by-step explanation:

A given quadratic has its vertex at (4, 6) and the point (1, -3). We want to write the equation in vertex form of the quadratic.

Recall that vertex form is given by:

$\displaystyle f(x) = a(x-h) ^2 + k$

Where (h, k) is the vertex and a is the leading coefficient.

Since our vertex is at (4, 6), h = 4 and k = 6:

$\displaystyle f(x) = a(x - 4)^2 + 6$

To determine the leading coefficient, since we are given that (1, -3) is a point on the parabola, when x = 1, y = -3. Substitute and solve for a:

$\displaystyle \begin{aligned} (-3) & = a((1)-4)^2 + 6 \\ \\ -3 & = a(-3)^2 + 6 \\ \\ 9a & = -9 \\ \\ a & = -1 \end{aligned}$

Hence, the leading coefficient a is -1.

Then our equation in vertex form is:

$\displaystyle f(x) = -(x-4)^2 + 6$