yes!
The equation y = 4x^2 + 5 is a parabola
It is a parabola with its vertex at the origin and the arms going upwards.
The equation of the parabola ( y = 4x^2 ) can be rewritten in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex. Here, it is clear that the vertex is at the origin (0, 0) and ( 4p = 4 ), giving ( p = 1 ). The focus of the parabola is located at ( (h, k + p) ), so the focus is at the point ( (0, 1) ).
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
yes!
The equation y = 4x^2 + 5 is a parabola
It is a parabola with its vertex at the origin and the arms going upwards.
The equation of the parabola ( y = 4x^2 ) can be rewritten in the standard form ( y = 4p(x - h)^2 + k ), where ( (h, k) ) is the vertex. Here, it is clear that the vertex is at the origin (0, 0) and ( 4p = 4 ), giving ( p = 1 ). The focus of the parabola is located at ( (h, k + p) ), so the focus is at the point ( (0, 1) ).
Their noses are both at the origin, and they both open upward, but y=4x2 is a much skinnier parabola.
y=4x2+3x+8
I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
The equation does not represent that of a parabola.
First you need to solve for y. So write 4x2+y=16 so y=16-4x2 Now write f(x)=16-4x2
No. [ y = 4x2 ] is a quadratic equation.
It is an up parabola.