The quadratic function f(x) =-5x^2+2x-3 has a minimum at the vertex? true or false?

Answer

False

Step-by-step explanation:

You can find the X value of the vertex by the formula: b/(-2a)

In order to understand who is 'b' and 'a', you have to think about the generic quadratic function:

f(x)=ax^2 +bx+c

So, in our case, b=+2 and a=-5

b/(-2a)= 2/(-1)(-10)=1/5

Than, the first derivate of f(x) gives you the value of a stationary point. It could be a maximum or a minimum

f'(x)= -10x +2

f'(x)=0 when x=1/5

So the X coordinate is the same.

But you have to verify if it is (or not) a minimum.

f'(x) is positive for x<1/5 and negative for x>1/5

Thus, f(x) is growing for x<1/5 and it is decreasing for x>1/5

x=1/5 is the vertex, but it isn't a minimum

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363

irrational, cannot