At what two points do the graphs of y=2x^2-5x-12 and y=1/2x^2-3x+4 intersect? explain your reasoning.
They have the same y, and x.
set them equal
2x^2-5x-12=1/2 x^2-3x+4
combine terms, solve for x (both values), then compute y for each x.
when i combine the like terms i don't know what the next step is
set y = y
2x^2 - 5x - 12 = (1/2)x^2 - 3x + 4
I don't like fractions, so multiply each term by 2
4x^2 - 10x - 24 = x^2 - 6x + 8
3x^2 -4x - 32 = 0
(x - 4)(3x + 8) = 0
x = 4 or x = -8/3
if x = 4, y = 32 - 20 - 12 = 0
if x = -8/3
y = 128/9 +40/3 - 12 = 140/9
I could have subbed into the other equation to obtain the same y values, since we set them equal
two points are(4,0) and (-8/3, 140/9)
To find the points where the two graphs intersect, we need to set their equations equal to each other and solve for x.
Step 1: Set the two equations equal to each other:
2x^2 - 5x - 12 = (1/2)x^2 - 3x + 4
Step 2: Simplify the equation:
2x^2 - 5x - 12 - (1/2)x^2 + 3x - 4 = 0
(3/2)x^2 - 2x - 16 = 0
Step 3: Multiply both sides of the equation by 2 to eliminate the fractions:
3x^2 - 4x - 32 = 0
Step 4: Now, we can solve the quadratic equation by factoring, completing the square, or by using the quadratic formula. Let's use the quadratic formula.
The quadratic formula is: x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 3, b = -4, and c = -32.
Substituting these values into the quadratic formula, we get:
x = (-(-4) ± √((-4)^2 - 4(3)(-32))) / (2(3))
x = (4 ± √(16 + 384)) / 6
x = (4 ± √400) / 6
x = (4 ± 20) / 6
Step 5: Simplify the expression:
x1 = (4 + 20) / 6 = 24 / 6 = 4
x2 = (4 - 20) / 6 = -16 / 6 = -8/3
The two points where the graphs intersect are (4, y) and (-8/3, y) for any y-value obtained by substituting x into either equation.