Answer:
[tex]\displaystyle f(x) = -(x-4)^2 + 6[/tex]
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:
[tex]\displaystyle f(x) = a(x-h) ^2 + k[/tex]
Where (h, k) is the vertex and a is the leading coefficient.
Since our vertex is at (4, 6), h = 4 and k = 6:
[tex]\displaystyle f(x) = a(x - 4)^2 + 6[/tex]
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:
[tex]\displaystyle \begin{aligned} (-3) & = a((1)-4)^2 + 6 \\ \\ -3 & = a(-3)^2 + 6 \\ \\ 9a & = -9 \\ \\ a & = -1 \end{aligned}[/tex]
Hence, the leading coefficient a is -1.
Then our equation in vertex form is:
[tex]\displaystyle f(x) = -(x-4)^2 + 6[/tex]
