How can I find the point of intersection of the graphs x^2=y=4 and 2x-y=1 by hand?
I can do the problem using a calculator, but I don't know how to do it by hand. I tried using substitution, but I ending up getting x^2+2x-5=0. It would not factor evenly. Should I use the quadratic to find x and then plug that value in for x and solve for y?
Please tell me if I'm on the right track.
Thanks,
Cori
Yes, use the quadratic equation.
Sure, let's go through the steps to find the point of intersection between the two graphs x^2 + y = 4 and 2x - y = 1 by hand.
Step 1: Write the system of equations:
x^2 + y = 4 ---(1)
2x - y = 1 ---(2)
Step 2: Solve Equation (2) for y to express it in terms of x:
y = 2x - 1
Step 3: Substitute the expression for y from Equation (2) into Equation (1):
x^2 + (2x - 1) = 4
Step 4: Simplify the equation to a quadratic equation in standard form:
x^2 + 2x - 5 = 0
Step 5: Using the quadratic formula, solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = 2, and c = -5
Substituting the values, we get:
x = (-2 ± √(2^2 - 4(1)(-5))) / (2(1))
x = (-2 ± √(4 + 20)) / 2
x = (-2 ± √24) / 2
x = (-2 ± 2√6) / 2
x = -1 ± √6
So, x ≈ -1 + √6 and x ≈ -1 - √6 are the two possible values for x.
Step 6: Substitute the obtained values of x back into Equation (2) to find the corresponding y-values:
For x ≈ -1 + √6:
y = 2x - 1
y = 2(-1 + √6) - 1
y = -2 + 2√6 - 1
y ≈ 1 + 2√6
For x ≈ -1 - √6:
y = 2x - 1
y = 2(-1 - √6) - 1
y = -2 - 2√6 - 1
y ≈ -3 - 2√6
So, the two possible points of intersection are approximately (-1 + √6, 1 + 2√6) and (-1 - √6, -3 - 2√6).
In conclusion, you were on the right track by using the quadratic formula to find the values of x. Make sure to substitute those values back into the equation to find the corresponding y-values and obtain the coordinates of the points of intersection.