Math  /  Algebra

QuestionFor a parabola in general form, f(x)=ax2+bx+cf(x)=a x^{2}+b x+c, where a0a \neq 0. If a<0a<0 then the range of f(x)f(x) is

Studdy Solution

STEP 1

1. The function f(x)=ax2+bx+cf(x)=ax^2+bx+c represents a parabola.
2. The coefficient a0a \neq 0 ensures the function is a quadratic.
3. Since a<0a<0, the parabola opens downward.
4. The vertex of the parabola gives the maximum value of f(x)f(x) because it opens downward.
5. The range of a parabola that opens downward is from -\infty to the y-coordinate of the vertex.

STEP 2

1. Find the x-coordinate of the vertex of the parabola.
2. Calculate the y-coordinate of the vertex using the x-coordinate found in step 1.
3. Determine the range of the parabola based on the vertex.

STEP 3

The x-coordinate of the vertex of a parabola f(x)=ax2+bx+cf(x)=ax^2+bx+c is given by
xv=b2a x_v = -\frac{b}{2a}

STEP 4

Substitute xv=b2ax_v = -\frac{b}{2a} into the function f(x)f(x) to find the y-coordinate of the vertex, f(xv)f(x_v):
f(b2a)=a(b2a)2+b(b2a)+c f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c

STEP 5

Simplify the expression for f(b2a)f\left(-\frac{b}{2a}\right):
f(b2a)=a(b24a2)b22a+c f\left(-\frac{b}{2a}\right) = a\left(\frac{b^2}{4a^2}\right) - \frac{b^2}{2a} + c f(b2a)=ab24a2b22a+c f\left(-\frac{b}{2a}\right) = \frac{ab^2}{4a^2} - \frac{b^2}{2a} + c f(b2a)=b24ab22a+c f\left(-\frac{b}{2a}\right) = \frac{b^2}{4a} - \frac{b^2}{2a} + c f(b2a)=b24a2b24a+c f\left(-\frac{b}{2a}\right) = \frac{b^2}{4a} - \frac{2b^2}{4a} + c f(b2a)=b22b24a+c f\left(-\frac{b}{2a}\right) = \frac{b^2 - 2b^2}{4a} + c f(b2a)=b24a+c f\left(-\frac{b}{2a}\right) = \frac{-b^2}{4a} + c

STEP 6

Since the parabola opens downward (because a<0a<0), the range of f(x)f(x) is all values less than or equal to the y-coordinate of the vertex:
Range(f(x))=(,b24a+c] \text{Range}(f(x)) = \left( -\infty, \frac{-b^2}{4a} + c \right]

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