Answer:
Vertex of given quadratic function \(f(x)=5x^2+20x-16\) is (-2,-36)
Step-by-step explanation:
Given function \(f(x)=5x^2+20x-16\)
We have to find the vertex of the given function \(f(x)=5x^2+20x-16\).
The standard quadratic function is represented by \(f(x)=a(x-h)^2+k\) ,
Where (h,k) represents the vertex and a ≠ 0.
If a is positive, the graph opens upward, and if a is negative, then it opens downward.
We first write the given equation in standard form ,by using completing square,
We know \((a-b)^2=a^2-2ab+b^2\)
First taking 5 common from function, we have,
\(f(x)=5(x^2+4x-\frac{16}{5})\)
Comparing we have, a = x and
-2ab= +4x ⇒ -2b = 4 ⇒ b = -2
Add and subtract b\(b^2=4\) , we get,
\(f(x)=5(x^2+4x+4-4-\frac{16}{5})\)
On simplifying, we have,
\(f(x)=5((x+2)^2+\frac{-20-16}{5})\)
\(f(x)=5((x+2)^2+\frac{-36}{5})\)
We get, \(f(x)=5((x+2)^2)-36\)
On comapring with standard equtaion , we have h = -2 and k = -36
Thus, vertex of given quadratic function \(f(x)=5x^2+20x-16\) is (-2 ,-36)
