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Chapter 2: Problem 29

Write the quadratic function in standard form and sketch its graph. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). $$f(x)=-x^{2}+2 x+5$$

Short Answer

Expert verified
The standard form of function is \(f(x)=-1*(x-1)^2+6\). The vertex is \((1,6)\). The axis of symmetry is \(x=1\). The x-intercepts are \((1-\sqrt{6}, 0) and (1+\sqrt{6}, 0)\).

Step by step solution

01

Rewrite the Quadratic Function

The quadratic function should be translated into its standard form. Given \(f(x)=-x^{2}+2 x+5\), it is seen that this is in the form \(f(x)=ax^{2}+bx+c\). The standard form of the quadratic function is \(f(x)=a(x-h)^{2}+k\), where \((h,k)\) is the vertex of the parabola.
02

Find the Vertex and Axis of Symmetry

After reordering the function in the standard form, easily find \(h\) and \(k\) by following the formula \(h=-\frac{b}{2a}\) and \(k=f(h)\). Identify \(h\) as the value of \(x\) that gives the axis of symmetry.
03

Find the X-intercepts

The x-intercepts are values of \(x\) for which \(f(x) = 0\). To solve this, set \(f(x)=-x^{2}+2 x+5 = 0\) and solve for \(x\) by either factoring, completing the square or using the quadratic formula.
04

Sketch the Graph

With the values of the vertex, axis of symmetry, and x-intercepts, the graph of the quadratic function can be sketched. The vertex is the maximum or minimum point of the graph, the axis of symmetry is a vertical line passing through the vertex, and the graph intersects the x-axis at the x-intercepts.

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Most popular questions from this chapter

Write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form. \(f(x)=x^{4}-4 x^{3}+5 x^{2}-2 x-6\) (Hint: One factor is \(\left.x^{2}-2 x-2 .\right)\)

Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.) $$5,3-2 i$$

Find the rational zeros of the polynomial function. $$f(z)=z^{3}+\frac{11}{6} z^{2}-\frac{1}{2} z-\frac{1}{3}=\frac{1}{6}\left(6 z^{3}+11 z^{2}-3 z-2\right)$$

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$f(x)=16 x^{3}-20 x^{2}-4 x+15$$

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$h(x)=2 x^{4}-3 x+2$$
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