Question

How do you find the vertex and the intercepts for $y = - {x^2} + 9$ ?

Answer 1

Hint: In the equation above, we have an equation with two variables, and we are supposed to find the vertex and intercepts of the same. In order to do this, we need the value of both $x$ and $y$ . For that we need to assume the values of either of them as $0$ , in order to know their intercepts. With the help of standard vertex form for a parabola formula, we can find the vertex values, which is
$y = m{(x - a)^2} + b$ with vertex $(a,b)$

Complete step-by-step solution:
For an equation $y = - {x^2} + 9$ , we have to first find the vertex and then the intercepts, this can be done with the help of the following formulas,
Vertex: \[\left( {0,9} \right)\]
$Y$ -intercept: \[y = 9\]
$X$ -intercepts: \[x = - 3\] and \[x = + 3\]
Explanation:
We all know that the standard vertex form for a parabola is
$ = m{\left( {x - a} \right)^2} + b$
with vertex at $(a,b)$
Re-writing the given equation: $y = - {x^2} + 9$ into explicit standard vertex form:
\[ \Rightarrow - 1{\left( {x - 0} \right)^2} + 9\]
with vertex at $(0,9)$
The $Y$ -intercept is the value of $y$ when $x = 0$
$ \Rightarrow {0^2} + 9$
Adding the numbers,
$ \Rightarrow 9$
The $X$-intercepts are the values of $x$ possible when $y = 0$
$ \Rightarrow - {x^2} + 9$
Shifting the variable to the other side,
$ \Rightarrow {x^2} = 9$
Taking square roots on both the sides,
$ \Rightarrow x = \pm 3$
The number will either be positive or negative after taking out the square root.

Therefore for the given question Vertex: \[\left( {0,9} \right)\]
$Y$ -intercept: \[y = 9\]
$X$ -intercepts: \[x = - 3\] and \[x = + 3\].

Note: The vertex form that we write is actually another form of writing out the equation of a parabola. The standard quadratic form is not really helpful while finding the vertex of a parabola. It needs to be converted to vertex form for that. For intercepts, $x$ -intercept is the point where the line crosses the $x$ -axis, and at this point $y = 0$ . The $y$ -intercept is the point where the line crosses $y$ -axis, and here $x = 0$ .