Question

How do you find the vertex and the intercepts for \[y = - {x^2} - 2x + 3\]?

Answer 1

Hint:
We can find the vertex and intercepts of the equation by their general formulas obtained and to find the y-intercept of the given equation, just we need to substitute \[x\]= 0 in the given equation and solve for y and to find the x-intercept of the given equation, just we need to substitute y = 0 in the given equation and solve for x.

Complete step by step solution:
The given equation is
\[y = - {x^2} - 2x + 3\]
The given equation is in the form of \[a{x^2} + bx + c\], in which we need to find the vertex of x and y coordinate.
Let us find vertex of x-coordinate
\[x = - \dfrac{b}{{2a}}\]
From the equation we know that it is quadratic equation and according to it, a = -1 and b = -2
Hence, we get
\[x = - \dfrac{{ - 2}}{{2\left( { - 1} \right)}}\]
Hence, the vertex of x-coordinate is
\[x = \dfrac{2}{{ - 2}} = - 1\]
\[x = - 1\].
Now let us find vertex of y-coordinate
As we obtained the value of x as -1, put this value in given equation to find y-coordinate as
\[y = - {x^2} - 2x + 3\]
\[y\left( { - 1} \right) = {\left( { - 1} \right)^2} - 2\left( { - 1} \right) + 3\]
Simplifying the terms, we get
\[y\left( { - 1} \right) = - 1 + 2 + 3\]
\[y\left( { - 1} \right) = 4\]
Therefore, the vertex obtained is \[\left( { - 1,4} \right)\].
We need to find the x and y intercepts for the equation given, hence to find the x-intercept substitute y=0 and solve for x using quadratic formula
\[y = - {x^2} - 2x + 3\]
\[ - {x^2} - 2x + 3 = 0\]
\[x = \dfrac{c}{a}\]
\[x = \dfrac{3}{{ - 1}}\]
\[x = - 3\]
Therefore, x intercepts at x = 1 and x = -3.
To find the y-intercept substitute x=0 and solve for y as
\[y = - {x^2} - 2x + 3\]
Implies that
y = 3.

Note:
As per the given equation consists of x and y terms based on the intercept asked, we need to solve for it. For ex if y-intercept is asked substitute x=0 and solve for y and if x-intercept is asked substitute y=0 and solve for x and the y-intercept of an equation is a point where the graph of the equation intersects the y-axis.