Interpreting Quadratic Functions in Standard Form

Quadratic functions can be written in standard form. $y = a x^{2} + b x + c$ We know that $a$ determines the width and if the graph is upward or downward, and $c$ gives the $y$ -intercept. Let's begin with the graphs of $f (x)$ and $h (x)$ , which have a positive value for the coefficient $a$ .

Positive $a$

The functions $B,$ $C$ and $E$ have positive coefficients in front of $x^{2}$ . $B . y = 0.6 x^{2} + 3 x + 3 C . y = x^{2} - 5 x - 10 E . y = 1.1 x^{2} - 2.5 x + 3$ We see directly that we can discard $C$ since the two graphs have vertical intercepts for $y = 3$ and not for $y = - 10$ . Further, $f (x)$ is somewhat wider than $h (x)$ , which in this case means that the value of $a$ is smaller. Since $0.6$ is less than $1.1$ , we can determine how the expressions fits the graphs. $B . f (x) y = 0.6 x^{2} + 3 x + 3 E . h (x) y = 1.1 x^{2} - 2.5 x + 3$

Negative $a$

The functions $A,$ $D$ and $F$ has negative coefficients in front of $x^{2}$ . We are to pair $g (x)$ and $k (x)$ with two of these: $A . y = - x^{2} + 7 x - 10 D . y = - 2 x^{2} + 5 x + 3 F . y = - 4 x^{2} - 12 x - 11 .$ We can discard $D$ since the $y$ -intercept is clearly below $3 .$ Another thing, observe that $k (x)$ is much wider than $g (x)$ , which means that it should have a less negative $a$ -value than $g (x)$ . Then you can see that $k (x)$ must be associated with $A$ and $g (x)$ with $F$ .

Summary

Thus, we have paired the graphs and functions in the following manner. $A . k (x) y = - x^{2} + 7 x - 10 B . f (x) y = 0.6 x^{2} + 3 x + 3 E . h (x) y = 1.1 x^{2} - 2.5 x + 3 F . g (x) y = - 4 x^{2} - 12 x - 11 .$