Question

Find the vertex of the function y = (-1/5)x2 + (4/5)x + 1/5.

Answer 1

ANSWER:

(2, 1)

EXPLANATION:

Given:

$-\frac{1}{5}x^2+\frac{4}{5}x+\frac{1}{5.}$

To find:

The vertex of the function

Step-by-step solution:

To be able to determine the vertex of the given function, we have to rewrite the function in the below vertex form;

$y=a(x-h)^2+k$

where (h, k ) is the vertex of the function.

Recall that a quadratic equation in general form is given as;

$y=ax^2+bx+c$

If we compare the above equation with the given equation, we see that a = -1/5, b = 4/5, and c = 1/5

We'll find the value of h using the below formula;

$\begin{gathered} h=-\frac{b}{2a} \\ h=-\frac{\frac{4}{5}}{2(-\frac{1}{5})} \\ h=\frac{-\frac{4}{5}}{-\frac{2}{5}} \\ h=-\frac{4}{5}*(-\frac{5}{2}) \\ h=\frac{4}{2} \\ h=2 \end{gathered}$

We can now determine the value of k as seen below;

$\begin{gathered} k=f(h) \\ \\ k=-\frac{1}{5}(2)^2+\frac{4}{5}(2)+\frac{1}{5} \\ \\ k=-\frac{4}{5}+\frac{8}{5}+\frac{1}{5} \\ \\ k=\frac{-4+8+1}{5} \\ \\ k=\frac{5}{5} \\ \\ k=1 \end{gathered}$

Since h = 2 and k = 1, then the vertex of the function is (2, 1) and we can write the function in vertex form as;

$y=-\frac{1}{5}(x-2)^2+1$