Video Transcript
Which of the following equations
could be represented by the given graph? Option (A) π₯ squared plus 10π₯
minus 24 equals zero. Option (B) negative π₯ squared
minus 24π₯ plus 10 equals zero. Option (C) π₯ squared minus 10π₯
minus 24 equals zero. Option (D) π₯ squared plus 10π₯
plus 24 equals zero. Or is it option (E) π₯ squared
minus 10π₯ plus 24 equals zero?
In this question, we are given five
different equations. And we need to determine which of
these equations can be represented by the given graph. To do this, we can start by
analyzing the graph we are given. We can see that the graph has a
parabolic shape and that we are given the values of its two π₯-intercepts, four and
six.
At this point, we might be tempted
to note that the parabola opens upwards. So the leading coefficient of the
quadratic must be positive. However, this is only true when
plotting graphs of quadratics. In this case, we are plotting the
graphs of equations. For instance, we cannot eliminate
option (B), since we could multiply both sides of the equation by negative one to
obtain the equation π₯ squared plus 24π₯ minus 10 equals zero. This is an equivalent equation; all
of its solutions are the same. However, the graph of the quadratic
on the left-hand side will be reflected in the π₯-axis. So we cannot use the leading
coefficient to eliminate options in this case.
Instead, we can recall that when
plotting the graph of a function, the π¦-coordinate of a point on the curve tells us
the output value of the function and the π₯-coordinate tells us the input value of
the function. In particular, this means that if
we were to sketch the graphs of these quadratic functions, then the points on the
graph whose π¦-coordinates are zero must be roots of the function and solutions to
the equation. Therefore, for an equation to be
represented by this graph, it must have solutions of π₯ equals six and π₯ equals
four.
So, we can check which of the five
given equations has solutions of π₯ equals four and π₯ equals six by
substitution. Letβs start with option (A). We can substitute π₯ equals four
into the left-hand side of the equation to obtain four squared plus 10 times four
minus 24. If we evaluate this expression,
then we see it is equal to 32. Thus, four is not a solution of
this equation, so it cannot be represented by the given graph.
We can follow the exact same
process for option (B). We substitute π₯ equals four into
the quadratic on the left-hand side of the equation and evaluate to get negative
102. Since this is nonzero, we can
conclude that four is not a root of this quadratic, so the graph cannot represent
this equation.
We can follow the same process for
option (C). We substitute π₯ equals four into
the quadratic on the left-hand side of the equation and evaluate to obtain negative
48. Since this is nonzero, we can
conclude that the graph cannot represent this equation.
Following the same process for
option (D) yields that the output of the quadratic at π₯ equals four is 80. Since the output is not zero at
this point, we can conclude that the graph cannot represent this equation.
For due diligence, we can also
apply this process to option (E). If we substitute π₯ equals four
into the quadratic on the left-hand side of the equation and evaluate, we obtain
zero. So π₯ equals four is a solution to
this equation. Similarly, if we substitute π₯
equals six into the quadratic on the left-hand side of this equation and evaluate,
we obtain zero. So π₯ equals six is also a solution
of this equation. Therefore, only option (E) has the
two roots given by the graph. We can also note that the equation
is a quadratic in π₯. So we would expect the shape of its
graph to be a parabola.
Hence, we can conclude that of the
five given options, only option (E), π₯ squared minus 10π₯ plus 24 equals zero,
could be represented by the graph.
