Question

what is the largest function value of f(x) = -x^2+4?

Answer 1

hint: look at the y coordinate of the vertex

Answer 2

how is it sir, i dont know how to solve?

Answer 3

do you know how to graph?

Answer 4

I would use a graphing tool like desmos

https://www.desmos.com/calculator

Answer 5

yes sir

Answer 6

how to solve this one sir, please help me?

Answer 7

ok graph it and see where the peak point is

Answer 8

type in `-x^2+4`

Answer 9

the graph is on maximum pick sir

Answer 10

because a is negative that is why the graph is downward

Answer 11

what's the highest point you see?

Answer 12

4 sir

Answer 13

(0,4) is the point, but y = 4 is the answer, yes

4 is the largest output possible

Answer 14

sir can you give me tips, how to analyze question easily?

Answer 15

you can think of y = -x^2 + 4 as y = -1(x-0)^2 + 4

then compare that to vertex form y = a(x-h)^2 + k

vertex = (h,k) = (0,4)

Answer 16

The general standard quadratic is the parabola shape. the equation of standard form is
y = ax^2 + bx + c

you were right with the first term, ax^2, and the sign determins if the parabola opens upwards or downwards in the y direction.

A parabola like this will have either a max value , or a min value depending which way it opens. For this prob there is a negative -ax^2, and it opens downwards. So it goes downwards forever to infinity, and so no minimum value. The maximum value will be at the vertex of the parabola.

Answer 17

You have

\(f(x) = -x^2+4\),

and you want to find the maximum value of f(x).

Notice you have x^2. No matter what value you use for x, x^2 will always be non-negative.
x^2 will have a minimum value of zero. x^2 is zero or greater.

The expression has -x^2, so -x^2 will be non-positive.
The maximum value of -x^2 is zero. -x^2 will be zero or lower.
Since the maximum value of -x^2 is zero, the maximum value of -x^2 + 4 is 4.