Is the vertex of the graph y=3(x+9)^2+5 a minimum value?

Question

sorry its y=3(x+9)^(2)+5

Answer 1

The minimum value of the graph is the y-coordinate of the vertex. It is not the vertex, but the value of the y-coordinate. In your case, the vertex is (-9, 5). Therefore, the minimum value is 5. That means that for every value of x not equal to -9 the value of the function will be greater than 5.

Answer 2

For any quadratic, write it indoors the style f(x) = a x² + b x + c, the situation a, b, & c are any actual numbers. Then the line of symmetry might desire to be x = -b / (2a). subsequently f(x) = -x² + 6 might have the line of symmetry at x = 0 / (-2) = 0, the y-axis. So x = 0. the utmost fee for this social gathering is definitely (0,6). Subtitute x = 0 into the equation f (0) - 6- 0 = 0. How do i are wakeful of it particularly is a optimal fee? i'm chuffed you asked! The learn in front of the x² term shows wherein course the graph faces. no count quantity if it particularly is effective, the graph is concave up, damaging the concave down.

Answer 3

Just an added note as to how the answer is calculated:

For parabolas, taking the derivative of this function gives it's slope function. Setting this slope = 0 gives the minimum or maximum for a parabola, depending if it opens up (positive slope) or down (negative slope).
This is positive, so the function has a minimum.

The derivative of this function is:
y' = 6x + 54,
when y' = 0, x = x = -9

The function has a minimum at (-9,5)

Answer 4

ywill be minimum when x=-9 when y=5

Answer 5

it is y=3(x+9)^(2+5) orrr y=3(x+9)^(2)+5


he's right... the minimum & vertex is (-9, 5)

Answer 6

Yes, because a quadratic equation of the form y = ax^2 + bx + c is always a parabola.