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The minimum or maximum of a quadratic function is the vertex on the parabola. Remember, quadratic functions (functions with x^2 is the term with the highest exponent) look like a U-shape, or parabola when you graph them. So you need to find the highest or lowest point of the U (depending on which way the U is going--up or down).
A vertex is a point, so you need to find the x and y for the point (x,y).
The formula or process for finding the vertex of a parabola is ( -b/(2a) , "plug in what you got for x into the quadratic equation"). You get b and a from the quadratic equation, which is written in the form y = ax^2 + bx + c.
Your equation is f(x)=-2x²+7x-3
So a = -2 and b = 7.
So the x-value of the vertex is -b/2a = -7/(2*-2) = -7/-4 = 7/4
To get the y-value of the vertex, plug in what you just got for x, 7/4, into the original equation:
Remember, you can replace f(x) with y.
y = -2x² + 7x - 3
y = -2(7/4)^2 + 7(7/4) - 3
Use PEMDAS to simplify.
y = -2(49/16) + 49/4 - 3
y = -49/8 + 49/4 - 3
y = -49/8 + 98/8 - 24/8
y = 49/8 - 24/8
y = 25/8
So the vertex of the parabola is (7/4 , 25/8).
This vertex is actually a maximum, because the parabola opens down since the a-value in the equation is negative (remember, a=-2).
When you find the domain of a function, you find all of the numbers you can plug into the equation for x. In this case (and it's the case for ALL quadratic functions), you can plug any real number in for x. So the domain is all real numbers.
The range represents all of the y-values you could get out of the function after plugging all possible numbers in for x. You can see it on the graph. The parabola in this problem, remember, opens DOWN, so all of your y-values are going to either be 25/8 (the value you got for the y on the vertex), or below that number. So the range is y<=25/8. In other words, y could never be higher than 25/8, because 25/8 is the maximum y-value for the function.I hope this helps you move forward on the problem! Let me know if you need any more clarification!
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