Find the max/min value of the function: f(x) = 5x^2 - 20x + 15

A quadratic function will either have a minimum or a maximum value depending on the shape of the function. A quadratic function with a positive value in front of the x^2 term has a U shape and therefore has a defined minimum value at the turning point of the graph. A function with a negative value in front of the x^2 term has an n shape and therefore has a defined maximum value at the turning point of the graph. The function in this question is therefore U shaped and has a defined negative value at the turning point.To find the minimum value of the quadratic, we must find at what point the gradient (or the slope) of the function is 0. At this point, the function is at its turning point (the minimum point on the graph). The gradient is found by differentiating the function. Therefore, the minimum value is found by differentiating the function (to find the gradient) and setting it equal to 0. Our function is f(x) = 5x^2 - 20x + 15. Applying simple differentiation rules (Multiplying the power by the coefficient and then reducing the power by 1) we get the follow term: f’(x) = 10x - 20. If we set that equal to zero we get: 10x - 20 = 0. Solving this equation gives us: x = 2. Therefore, at x = 2, the quadratic has a minimum value. To find the y value of the minimum, we simply sub x = 2 back into the original equation. y = 5(2)^2 - 20(2) +15 gives us a y value of -5. The minimum point is therefore at (2,-5) 

Answered by Peter H. Maths tutor

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