Which function has a vertex at the origin? f(x) = (x + 4)2 f(x) = x(x – 4) f(x) = (x – 4)(x + 4) f(x) = –x2
Answer: \(f(x)=-x^2\)
Step-by-step explanation:
In mathematics,the starting point on a grid is called an origin. It is the point (0,0), where the x-axis and y-axis intercept.
Thus to check which function passing through the origin, we need to check for x=0, y=f(x) must equals to 0.
1. \(f(x)=(x+4)^2\)
At x=0,\(f(0)=(0+4)^2=4^2=16\)
⇒f(x) is not passing through origin.
2. \(f(x)=x(x-4)\)
At x=0,\(f(0)=0(0-4)=0\)
⇒f(x) passing through origin.
But the vertex form of equation is \(f(x)=x^2-4x-4+4=(x-2)^2-4\)
⇒ vertex of f(x)=(2,4)
3. \(f(x)=(x-4)(x+4)\)
At x=0,\(f(0)=(0-4)(0+4)=-4\times4=-16\)
⇒f(x) is not passing through origin.
4. \(f(x)=-x^2\)
At x=0,\(f(0)=0^2=0\)
⇒f(x) passing through origin.
Vertex form of equation=\(f(x)=-1(x-0)+0\)
⇒ vertex=(0,0)
Answer:
f(x) = –x2
Step-by-step explanation:
i just did the test and got it right