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The vertex of this parabola is at 2 -4 When the y-value is -3 the x-value is -3 What is the coefficient of the squared term in the parabola's equation?
The general equation of a parabola is y = ax2 + bx + c.The vertex of a parabola is (-b/2a, c - b2/4a). In our case the vertex is (2, -4). So we have,-b/2a = 2, so that b = -4a, andc - b2/4a = -4 (substitute -4a for b)c - (-4a)2/4a = -4c = 16a2/4a = -4 (simplify)c - 4a = -4 (solve for c)c = 4a - 4By substituting -4a for b, and 4a - 4 for c, the equation becomes,y = ax2 + bx + cy = ax2 + (-4a)x + 4a - 4Since it is given that when x = -3, y = -3, we substitute them into the new equation and solve it for a. So we have,y = ax2 + (-4a)x + 4a - 4-3 = a(-3)2 + (-4a)(-3) + 4a - 4-3 = 9a + 12a + 4a - 4 (add 4 to both side, and add alike terms on the right-hand side)1 = 25a (divide by 25 to both sides)1/25 = a (this is the required answer)If you want to find the equation of the parabola, just substitute 1/25 for a, such as:y = ax2 + (-4a)x + 4a - 4y = (1/25)x2 + [-4(1/25)]x + 4(1/25) - 4y = (1/25)x2 -4/25x + 4/25 - 4y = (1/25)x2 -4/25x + 96/25
What is the order pair for 4a - 5b equals 7 and 4a plus 5b equals 17?
Simultaneous Equations 4a - 5b = 7 4a + 5b = 17 Add the two , this will eliminate 'b' 8a = 24 Divide both sides by '8' a = 3 When a= 3 , substitute into eityher equation for 'b'. 4(3) + 5b = 17 12 + 5b = 17 5b = 17 - 12 = 5 5b = Divide both sides by '5'. Hence 'b = 1'.
How do you find the derivative of a function that has been raised to a power greater than 1?
Let us say that f(x)=x^4A derivative is the opposite to an integral.If you were to integrate x^4, the first process is taking the power [which in this case is 4], multiplying it by any value before the x [which is 1], then subtracting 1 from the initial power [4]. This leaves 4x^3. The final step is taking the integral of what is 'inside' the power [which is (x)], and multiplying this to the entire answer, which results in 4x^3 x 1 = 4x^3If you were to derive (x)^4, you would just add 1 to the power [4] to become (x)^5 then put the value of the power as the denominator and the function as a numerator. This leaves [(x^5)/(5)]To assure that the derivative is correct, integrate it. (x^5) would become 5x^4. Since (x^5) is over (5), [(5x^4)/(5)] cancels the 5 on the numerator and denominator, thus leaving the original function of x^4
