5.1 Quadratic functions  (Page 2/15)

The standard form of a quadratic function presents the function in the form

where $\left(h,k\right)$ is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function    .

As with the general form, if $a>0,$ the parabola opens upward and the vertex is a minimum. If $a<0,$ the parabola opens downward, and the vertex is a maximum. [link] represents the graph of the quadratic function written in standard form as $y=\mathrm{-3}{\left(x+2\right)}^2+4.$ Since $x–h=x+2$ in this example, $h=\mathrm{–2.}$ In this form, $a=\mathrm{-3},h=\mathrm{-2},$ and $k=4.$ Because $a<0,$ the parabola opens downward. The vertex is at $\left(-2,\text{ 4}\right).$

The standard form is useful for determining how the graph is transformed from the graph of $y=x^2.$ [link] is the graph of this basic function.

If $k>0,$ the graph shifts upward, whereas if $k<0,$ the graph shifts downward. In [link] , $k>0,$ so the graph is shifted 4 units upward. If $h>0,$ the graph shifts toward the right and if $h<0,$ the graph shifts to the left. In [link] , $h<0,$ so the graph is shifted 2 units to the left. The magnitude of $a$ indicates the stretch of the graph. If $\left|a\right|>1,$ the point associated with a particular $x\text{-}$ value shifts farther from the x- axis, so the graph appears to become narrower, and there is a vertical stretch. But if $\left|a\right|<1,$ the point associated with a particular $x\text{-}$ value shifts closer to the x- axis, so the graph appears to become wider, but in fact there is a vertical compression. In [link] , $\left|a\right|>1,$ so the graph becomes narrower.

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

For the linear terms to be equal, the coefficients must be equal.

This is the axis of symmetry    we defined earlier. Setting the constant terms equal:

In practice, though, it is usually easier to remember that k is the output value of the function when the input is $h,$ so $f(h)=k.$

Writing the equation of a quadratic function from the graph

Write an equation for the quadratic function $g$ in [link] as a transformation of $f(x)=x^2,$ and then expand the formula, and simplify terms to write the equation in general form.

We can see the graph of g is the graph of $f(x)=x^2$ shifted to the left 2 and down 3, giving a formula in the form $g(x)=a{(x-(\mathrm{-2}))}^2-3=a{(x+2)}^2–3.$

Substituting the coordinates of a point on the curve, such as $(0,\mathrm{-1}),$ we can solve for the stretch factor.

$$\begin{array}{ccc}\hfill -1& =& a{\left(0+2\right)}^2-3\hfill \\ \hfill 2& =& 4a\hfill \\ \hfill a& =& \frac{1}{2}\hfill \end{array}$$

In standard form, the algebraic model for this graph is $(g)x=\frac{1}{2}{(x+2)}^2–3.$

To write this in general polynomial form, we can expand the formula and simplify terms.

$$\begin{array}{ccc}\hfill g(x)& =& \frac{1}{2}{(x+2)}^2-3\hfill \\ & =& \frac{1}{2}(x+2)(x+2)-3\hfill \\ & =& \frac{1}{2}(x^2+4x+4)-3\hfill \\ & =& \frac{1}{2}{x}^2+2x+2-3\hfill \\ & =& \frac{1}{2}{x}^2+2x-1\hfill \end{array}$$

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

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