State the coordinates of the vertex and the number of $x$-intercepts for the following function: $$ y = -4x^2 + 1 $$
I am not really asking for a straight-up answer. If you could please tell me HOW you found the vertex and the number of $x$-intercepts (or how many times it intersects the $x$-axis) that would be really great, so I could do the rest of the questions.
On
You can note that since the quadratic is just a stretch of factor $-4$ units and followed by a vertical translation of $1$ unit of the graph $y=x^2$ the vertex will be found at $$\left(0,1 - 4\cdot 0^2\right) = \left(0,1\right)$$
To see how many times it intercepts the $x$-axis or intercepts the line $y=0$, we need only set $y=0$ and solve to see that our quadratic factorises as $$0 = -(2x-1)(2x+1) \iff x = \pm \frac{1}{2}$$ which is easily seen since $1 -(2x)^2 = 1-4x^2$ is a difference of two squares. So there are two solutions.
An alternative is to compute the determinant of the quadratic, namely $$\Delta = 0 - 4(-4)(1) = 16 > 0$$ since the determinant is positive, then the quadratic has two roots or has two $x$-intercepts.
NB: The determinant of a quadratic $ax^2 + bx + c$ is $\Delta = b^2 - 4ac$
The vertical parabola with vertex at $(h,k)$ can be written as
$$y-k=a(x-h)^2$$
for some non-zero value $a$. So to find the vertex of your parabola, rewrite your given equation in that form. Your first step should be to move that $+1$ on the right-hand side of your equation to the left-hand side. Then continue from there to find what $h$ and $k$ are.
@GTonyJacobs explains how to find the number of $x$-intercepts. At any $x$-intercept, the $y$-coordinate is zero. So substitute $y=0$ into your equation and you will get a quadratic equation with only one variable, $x$. Solve that equation for $x$ in any of the standard ways (square root of both sides, graphing, factoring, completing the square, quadratic formula) and see how many real values of $x$ you get. You will get no, one, or two real values of $x$. See which one you get in your particular case.