Video Transcript
Solve 𝑥 squared minus 𝑥 minus six
equals zero by factoring, and hence determine which of the following figures would
be a sketch of 𝑦 is equal to 𝑥 squared minus 𝑥 minus six. Is it figure (A), (B), (C), (D), or
(E)?
The first part of the question
asked us to solve the quadratic equation 𝑥 squared minus 𝑥 minus six equals zero
by factoring. Since the coefficient of 𝑥 squared
is one, we know that we can rewrite the quadratic in the form 𝑥 squared plus 𝑝
plus 𝑞 𝑥 plus 𝑝𝑞, which in turn can be factored into two sets of parentheses 𝑥
plus 𝑝 and 𝑥 plus 𝑞.
Since the coefficient of 𝑥 in our
equation is negative one, we know that 𝑝 plus 𝑞 is equal to negative one. Likewise, since the constant in our
equation is negative six, 𝑝𝑞 equals negative six. Hence, we need to find two value
that have a sum of negative one and a product of negative six. Negative three plus two equals
negative one. And negative three multiplied by
two equals negative six. This means that 𝑥 squared minus 𝑥
minus six equals zero can be rewritten as 𝑥 minus three multiplied by 𝑥 plus two
equals zero.
Since the product of our two
factors equals zero, then one of the factors must itself equal zero. So 𝑥 minus three equals zero or 𝑥
plus two equals zero. This means that either 𝑥 equals
three or 𝑥 equals negative two. The two solutions to the quadratic
equation 𝑥 squared minus 𝑥 minus six equals zero are 𝑥 equals three and 𝑥 equals
negative two.
The second part of the question
asked us to identify which of the graphs is a sketch of the equation 𝑦 is equal to
𝑥 squared minus 𝑥 minus six. Recalling that the roots of a
function, which are the same as the solutions of the corresponding equation, tell us
what values of 𝑥 give us 𝑦 equals zero, we know that the curve will cross the
𝑥-axis at three and negative two. This is only true of graph (E).
Additionally, let us note other
aspects of the graph that we can use to verify the correct answer. Firstly, the 𝑦-intercept of the
graph is negative six, which corresponds to the fact that any quadratic function of
the form 𝑦 is equal to 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 will have a 𝑦-intercept at
𝑐. Secondly, since 𝑎 is greater than
zero, we know that the graph must open upwards, which also happens in this graph and
would allow us to eliminate options (A) and (C).
We can therefore conclude that the
figure that would be a sketch of 𝑦 is equal to 𝑥 squared minus 𝑥 minus six is
graph (E).
