{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} Quadratic functions can be written in standard form. $y=ax_{2}+bx+c$ 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 $a$.

The functions $B,$ $C$ and $E$ have positive coefficients in front of $x_{2}$. $ B.y=0.6x_{2}+3x+3C.y=x_{2}−5x−10E.y=1.1x_{2}−2.5x+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=0.6x_{2}+3x+3E.h(x)y=1.1x_{2}−2.5x+3 $

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}+7x−10D.y=-2x_{2}+5x+3F.y=-4x_{2}−12x−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 $F$.

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