# How do you find the range of #f(x)=5x^2+2x-1#?

Range:

graph{[-2, 2, -5, 5]} = 5x^2 + 2x-1

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To find the range of the function f(x) = 5x^2 + 2x - 1, you can use calculus techniques or observe the properties of quadratic functions. Since the leading coefficient of the quadratic term is positive (5), the parabola opens upwards. This means the range of the function is all real numbers greater than or equal to the vertex's y-coordinate. To find the vertex, use the formula x = -b/2a. Substituting the coefficients from the function, x = -2/(2*5) = -1/5. Then, plug this x-value into the function to find the y-coordinate of the vertex: f(-1/5) = 5(-1/5)^2 + 2(-1/5) - 1 = -1/5. Therefore, the vertex is (-1/5, -1/5). Since the parabola opens upwards, the range is all real numbers greater than or equal to the y-coordinate of the vertex. So, the range of f(x) = 5x^2 + 2x - 1 is [-1/5, ∞).

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To find the range of ( f(x) = 5x^2 + 2x - 1 ), you need to determine the possible values that ( f(x) ) can take. Since ( f(x) ) is a quadratic function, its range depends on the vertex of the parabola it represents.

The range of ( f(x) ) can be found using the formula for the vertex of a quadratic function, which is ( \frac{-b}{2a} ) for a quadratic function in the form ( ax^2 + bx + c ), where ( a ), ( b ), and ( c ) are coefficients.

In this case, ( a = 5 ) and ( b = 2 ). Therefore, the x-coordinate of the vertex is ( \frac{-2}{2(5)} = -\frac{1}{5} ).

Substitute this x-coordinate into the function to find the corresponding y-coordinate:

( f\left(-\frac{1}{5}\right) = 5\left(-\frac{1}{5}\right)^2 + 2\left(-\frac{1}{5}\right) - 1 )

Simplify this expression to find the y-coordinate of the vertex, which will give you the minimum or maximum value of the function depending on the sign of the coefficient of the squared term.

After finding the vertex, determine whether the parabola opens upwards or downwards. This will tell you whether the range is bounded below or above.

In this case, since the coefficient of the squared term (5) is positive, the parabola opens upwards, so the minimum value occurs at the vertex.

Thus, the range of ( f(x) ) is all real numbers greater than or equal to the y-coordinate of the vertex.

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