Question

How do you find the vertex and intercepts for $y = {(x + 3)^2} - 4$?

Answer 1

Here in this question, we have to determine the vertex and intercepts for the given equation. By substituting the x is equal to zero we can find the y intercept value and substituting y is equal to zero we can find the x intercept value. By equating the given equation to the general vertex equation we can determine the vertex.

Complete step-by-step solution:
The given equation is in the form of $f(x) = a{(x - h)^2} + k$, where (h, k) represents the vertex for the equation. An intercept is a point where the straight line or a curve intersects the y-axis in a plane. If the point x is zero then the obtained point is a y -intercept.
Now consider the given equation $y = {(x + 3)^2} - 4$ -----------(1)
Substitute the value of x as 0 in equation (1), then we have
$\Rightarrow y = {(0 + 3)^2} - 4$
On simplifying we get
$\Rightarrow y = 9 - 4 = 5$
Therefore, y-intercept is (0,5)
Substitute the value of y as 0 in the equation (1) then we have
$\Rightarrow 0 = {(x + 3)^2} - 4$
On simplifying we get
$\Rightarrow 4 = {(x + 3)^2}$
applying square root
$\Rightarrow \pm 2 = (x + 3)$
therefore we have
$\Rightarrow 2 = x + 3$ and $- 2 = x + 3$
$\Rightarrow x = - 1$ and $x = - 5$
Therefore, x-intercept is (-1, 0) and (-5, 0)
The general vertex equation of a line is given by$f(x) = a{(x - h)^2} + k$,----- (2) where (h, k) is a vertex
On comparing the equation (1) and equation (2) we get
$(h,k) = ( - 3, - 4)$
Therefore, the vertex is $( - 3, - 4)$

Note: The equation represents the equation of a line. The intercepts are the value obtained by equating the anyone parameter to zero and hence find the value. the vertex is nothing but the point. The vertex form of a quadratic is given by $f(x) = a{(x - h)^2} + k$ where (h, k) is the vertex. the f(x) can be assumed as y also.