Respuesta :
Answer:
(3, -16)
Step-by-step explanation:
The vertex coordinates of f(x) = ax² + bx + c are: (h, k)
where: [tex]h=\dfrac{-b}{2a}\ ,\qquad k=f(h)[/tex]
f(x) = x² - 6x + 11 ⇒ a = 1 b = -6
So:
[tex]h=\dfrac{-(-6)}{2(1)}=\dfrac6{2}=3\\\\y=f(h)=-(3)^2-6(3)+1=-9-18+11=-16[/tex]
The vertex of quadratic equation is (3, 2)
How to find the vertex of quadratic equation?
In order to determine the coordinates of the vertex, if they are not clearly labeled or easily identified from the graph, we will need to know the equation for the parabola. In general, the equation of a parabola has the form
y=ax²+bx+c
where a, b, and c are all real numbers, and a≠0.
This is called a quadratic equation, and this form is called the standard form of a quadratic equation.
The vertex of the parabola is located at a pair of coordinates which we will call (h, k). In order to determine what they are, we can simply use the standard form of the quadratic equation.
- To find h, use the formula:
h=−b/2a.
- To find k, evaluate the quadratic equation when x = h. In other words, plug in the value of h for x, and solve.
Given:
y= x² - 6x + 11
Now,
h= -b/2a
h= -(-6)/ 2 * 1
h= 3
Now,
y= f(h)
= 3*3 - 6*3 +11
= 9 -18 +11
= -9 +11
= 2
Learn more about vertex of equation here:
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