How do you use the important points to sketch the graph of #Y=x^2 - 16#?

Question

Eva Adamson · Accepted Answer

y-intercept = Vertex#->(x,y)=(0,16)#
x-intercept at point: #(4,0)" and "(-4,0)# Compare to #y=x^2# which is of shape #uu# and the vertex is at the point# (x,y)=(0,0)# Now include the #-16# and it drops the whole thing down by 16. So the vertex for #y=x^2-16# is at: #color(blue)("Vertex"->(x,y)=(0,-16)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ To determine the x-intercepts set #y=0# giving #y=0=(x^2-4^2) larr #always be on the lookout for this one #0=(x-4)(x+4)# Set #x-4=0 -> x=+4# Set #x+4=0->x=-4#

Isaiah Anthony · Answer

undefined To sketch the graph of $y = x^2 - 16$ using important points, we can identify key features such as the vertex, intercepts, and symmetry.

1. **Vertex**: The vertex of the parabola given by $y = x^2 - 16$ can be found using the formula $(-b/2a, f(-b/2a))$, where $a$ and $b$ are the coefficients of the quadratic equation. Here, $a = 1$ and $b = 0$, so the vertex is $(0, -16)$.

2. **Intercepts**: To find the $y$-intercept, set $x = 0$, which gives $y = 0^2 - 16 = -16$. So, the $y$-intercept is $(0, -16)$. To find the $x$-intercepts, set $y = 0$ and solve for $x$. $0 = x^2 - 16$ gives $x = \pm \sqrt{16} = \pm 4$. So, the $x$-intercepts are $(-4, 0)$ and $(4, 0)$.

3. **Symmetry**: Since the equation is $y = x^2 - 16$, the parabola is symmetric about the y-axis.

Using these points, we can sketch the graph:
- Plot the vertex at $(0, -16)$.
- Plot the $y$-intercept at $(0, -16)$.
- Plot the $x$-intercepts at $(-4, 0)$ and $(4, 0)$.
- Draw a smooth curve that passes through these points, remembering that the parabola opens upwards.

The resulting graph will be a parabola opening upwards, centered at the point $(0, -16)$, passing through the points $(-4, 0)$ and $(4, 0)$.