can you please show me in steps how to find f(1+a)

in this equation
f(x)=-2x^2+4x+6
i know that you will plug in 1+a anywhere there is x. but after that i get confused.

also can you tell me how to find the domain and range of a relation
for ex. 2x+5y=10
thanks for your help!

f(x)=-2x^2+4x+6

f(1+a) = -2(1+a)^2 + 4(1+a) + 6
= -2(1 + 2a + a^2) + 4 + 4a + 6
= -2 -4a -2a^2 + 10 + 4a
= -2a^2 + 8

for 2x+5y=10 both the domain and range are the real numbers
You are looking at a slanted line, (slope=-2/5), so you will find the graph for whatever x you use, and for whatever y you use.
This is true for all straight line graphs, which are neither vertical nor horizontal.

btw, why are you calling this "trig" ?

ok,thanks alot.. what about this one : y=2x^2-5

its in the first chapter of my trig course. i know its review frm geometry and alg

this is a vertical parabola, meaning it opens either upwards or downwards

the domain of any vertical parabola is the set of real numbers

for the range:
if the parabola opens up, then y ≥ the y value of the vertex
if the parabola opens down, then y ≤ the y value of the vertex

in this case
Domain : x any real number
Range : y ≥ -5, y being real

ok. so do u mean that to figure this out u always have to graph the relation?

no, you just have to find the vertex.

Here is a quick way to find the vertex:

if y = ax^2 + bx + c

then the x of the vertex is -b/(2a)
sub that back into the function to find the y value.

im sry im still kinda confused. can you show me once again please and explain everything you do thnks:

y=2x+3

y=4/square root of x^2+81

thnks alot!!

To find f(1+a) in the equation f(x) = -2x^2 + 4x + 6, follow these steps:

Step 1: Replace x with 1+a in the equation.
f(1+a) = -2(1+a)^2 + 4(1+a) + 6

Step 2: Simplify the expression.
f(1+a) = -2(1 + 2a + a^2) + 4 + 4a + 6
= -2 - 4a - 2a^2 + 4 + 4a + 6
= -2a^2 + 8

Therefore, f(1+a) = -2a^2 + 8.

Now, let's move on to finding the domain and range of the relation 2x + 5y = 10.

To find the domain, we need to determine the possible values for x. In this equation, x can take any real value since there are no restrictions on it. Therefore, the domain is all real numbers (-∞, ∞).

To find the range, we need to determine the possible values for y. We can rearrange the equation to isolate y:
2x + 5y = 10
5y = -2x + 10
y = (-2x + 10)/5

From this form, we can see that y depends on x. As x varies, y will also vary. Therefore, the range is also all real numbers (-∞, ∞).

In summary, the domain of the relation 2x + 5y = 10 is all real numbers, and the range is also all real numbers.