SOLUTION: Hello! I need some help with this equation.
Solve y = 2x^2 + 1 when x = -2.
I got y= -7, but am unsure if it is the correct answer. Any assistance would be greatly appreciat
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-> SOLUTION: Hello! I need some help with this equation.
Solve y = 2x^2 + 1 when x = -2.
I got y= -7, but am unsure if it is the correct answer. Any assistance would be greatly appreciat
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Question 1174347: Hello! I need some help with this equation.
Solve y = 2x^2 + 1 when x = -2.
I got y= -7, but am unsure if it is the correct answer. Any assistance would be greatly appreciated! Found 2 solutions by math_tutor2020, mccravyedwin:Answer by math_tutor2020(3816) (Show Source):
Replace every copy of x with -2 and evaluate to get
y = 2(x)^2 + 1
y = 2(-2)^2 + 1
y = 2(-2)*(-2) + 1
y = 2(4) + 1
y = 8 + 1
y = 9
I have a feeling your steps probably looked like this
y = 2x^2 + 1
y = 2(-2)^2 + 1
y = 2(-4) + 1
y = -8 + 1
y = -7
Which is not correct because (-2)^2 is equal to positive 4, and not negative 4.
Think of (-2)^2 as (-2)*(-2) = 4. The two negatives multiply to a positive.
In short, we're squaring all of the value and that includes the negative.
Substitute (-2) in place of x in
y = 2x^2 + 1
We get
y = 2(-2)^2 + 1
Substitute (4) in place of (-2)^2 because -2 times itself gives +4
y = 2(4) + 1
Substitute 8 in place of 2(4) because 2 times 4 is 8
y = 8 + 1
Substitute 9 in place of 8 + 1 because 8 plus 1 is 9
y = 9.
Oh, oh, you got -7. See if you can find out where you went wrong.
Edwin