Find the vertex and the axis of symmetry for the parabola with the given equation y=5x2-30+47....
I will assume you have a typo, with the first term 5x^2 and that the second term is -30x
the x of the vertex = -b/(2a) = 30/(10) = 3
sub in x=3
y = 5(9) - 30(3) + 47 = 2
vertex (3,2)
axis of symmetry : x = 3
Thanks, I was getting something like 9
now I see what I did wrong.
To find the vertex and axis of symmetry for the given parabola equation y = 5x^2 - 30x + 47, we can use the standard form of a quadratic equation, which is y = ax^2 + bx + c.
In this case, a = 5, b = -30, and c = 47.
The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
To find the vertex, we can use the following formulas:
h = -b/2a
k = f(h)
First, let's find the value of h:
h = -(-30) / 2*5
h = 30 / 10
h = 3
Now, let's find the value of k:
k = 5(3)^2 - 30(3) + 47
k = 5(9) - 90 + 47
k = 45 - 90 + 47
k = 2
Therefore, the vertex of the parabola is (3, 2).
The axis of symmetry is a vertical line that passes through the vertex. In this case, the axis of symmetry is the line x = 3.
So, the vertex of the parabola is (3, 2), and the axis of symmetry is x = 3.