Question

Graph the equation y=4^2+16x-3
1. Direction

2. Y intercept

3. Axis of Symmetry

4. Vertex

5. Zeros

Answer 1

so what direction is it going in first off

Answer 2

The equation was graphed correctly by @ch3wyc00ki3.

The direction is positive since it goes from left to right.

The Y-Intercept (otherwise where the line crosses the Y-axis) is at (0,13).

Not enough elaboration in terms of the vertex, you'd need to have shapes on the graph to find those, and here, as far as I can see it's simply a constant line with a positive slope.

The zero of a graph (and I assume it to be the real zero here) is basically just the x-intercept, or the opposite of the y-intercept. As such, at least on my graphing calclulator, your real zero should be 0.8125 or (0.8125, 0).


That's about as much as I can help you on with the information provided. If there's more to this and I have the entirely wrong idea then tell me, because not many details were provided, you just told us to graph the equation and listed some terms which I assume you want us to find.

Answer 3

\[y=4x^2+16x-3\]
\[y=4(x^2+4x+4-4)-3\]\[=4(x+2)^2-16-3\]\[=4(x+4)^2-19\]
1. it is an upward parabola.
2.y -intercept =-3
3.axis of symmetry is x+4=0 or x=-4
4. vertex : (-4,-19)
5.to find zeros put y=0
\[4x^2+16x-3=0\]
\[x=\frac{ -16 \pm \sqrt{16^2-4\times 4\times -3} }{ 2\times4 }\]
\[x=\frac{ -16 \pm \sqrt{256+48} }{ 8 }\]
\[x=\frac{ -16 \pm \sqrt{304} }{ 8}\]
\[x=\frac{ -16 \pm 4\sqrt{19} }{ 8 }\]
\[x=-2 \pm\frac{ \sqrt{19}}{2}\]

Answer 4

if you dont know the answer just look it up on google its way better than this bruh