Which is an example of a quadratic function whose graph opens downward and has its vertex at (0, 1)? A. y = −x^2 + 1
B. y = x^2 + x − 1
C. y = x^2 + 4x + 1
D. y = x^2 − 1

Given:

The graph of a quadratic function opens downward and has its vertex at (0, 1).

To find:

The quadratic function.

Solution:

The vertex form of a quadratic function is:

         ...(i)

Where, a is a constant and (h,k) is the vertex.

If a<0, then graph opens downward and if a>0, then graph opens upward.

It is given that the quadratic function opens downward and has its vertex at (0, 1). It means a must be negative.

Putting h=0 and k=1 in (i), we get

For ,

So, option A is correct.

In other options the leading coefficient is positive it means their graphs open upward. So, options B, C and D are incorrect.

Therefore, the correct option is A.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: