Let f(x)=2x^2-x+5 and g(x)=3x^2-3x-2 calculate h(1), if h(x)=f(x)+g(x)

I don't understand how I'm supposed to solve this. If anyone knows the answer or how to solve it it would be great.

h(1) = f(1)+g(1) = (2*1-1+5)+(3*1-3-2) = 6-2 = 4

or,

h(x) = f(x)+g(x)
= 2x^2-x+5 + 3x^2-3x-2
= 5x^2-4x+3
so, h(1) = 5-4+3 = 4

Oh okay I get it, thanks for explaining.

To calculate h(1), you need to substitute the value x=1 into the expression for h(x), which is h(x) = f(x) + g(x).

First, let's find the value of f(1):
f(x) = 2x^2 - x + 5
Substitute x = 1 into the expression for f(x):
f(1) = 2(1)^2 - 1 + 5
f(1) = 2(1) - 1 + 5
f(1) = 2 - 1 + 5
f(1) = 1 + 5
f(1) = 6

Next, let's find the value of g(1):
g(x) = 3x^2 - 3x - 2
Substitute x = 1 into the expression for g(x):
g(1) = 3(1)^2 - 3(1) - 2
g(1) = 3(1) - 3 - 2
g(1) = 3 - 3 - 2
g(1) = 0 - 2
g(1) = -2

Now that we have the values of f(1) = 6 and g(1) = -2, we can find h(1) by substituting these values into the expression for h(x):
h(x) = f(x) + g(x)
h(1) = f(1) + g(1)
h(1) = 6 + (-2)
h(1) = 6 - 2
h(1) = 4

Therefore, h(1) is equal to 4.