For the following quadratic functions, determine their shape, turning point, and x-intercepts: a) f(x) = x2, b) y = 2x2 − 2, c) 10x2 + 5y = -20

The coefficient of $x^2$ is $a=1$. Since $a > 0$, the parabola opens upwards.
Shape: Happy face

For $f(x) = x^2$, we have $a=1$, $b=0$, $c=0$.
The x-coordinate of the turning point is $x = -\frac{b}{2a} = -\frac{0}{2(1)} = 0$.
The y-coordinate is $f(0) = (0)^2 = 0$.
Turning point: $\boxed{(0, 0)}$

Set $f(x) = 0$:

x-intercept: $\boxed{(0, 0)}$

The coefficient of $x^2$ is $a=2$. Since $a > 0$, the parabola opens upwards.
Shape: Happy face

For $y = 2x^2 - 2$, we have $a=2$, $b=0$, $c=-2$.
The x-coordinate of the turning point is $x = -\frac{b}{2a} = -\frac{0}{2(2)} = 0$.
The y-coordinate is $y = 2(0)^2 - 2 = -2$.
Turning point: $\boxed{(0, -2)}$

Set $y = 0$:

x-intercepts: $\boxed{(-1, 0) \text{ and } (1, 0)}$

The coefficient of $x^2$ is $a=-2$. Since $a < 0$, the parabola opens downwards.
Shape: Sad face

For $y = -2x^2 - 4$, we have $a=-2$, $b=0$, $c=-4$.
The x-coordinate of the turning point is $x = -\frac{b}{2a} = -\frac{0}{2(-2)} = 0$.
The y-coordinate is $y = -2(0)^2 - 4 = -4$.
Turning point: $\boxed{(0, -4)}$

Set $y = 0$:

Since $x^2$ cannot be negative for real numbers, there are no real x-intercepts.
x-intercepts: $\boxed{\text{No real x-intercepts}}$

The coefficient of $x^2$ is $a=-4$. Since $a < 0$, the parabola opens downwards.
Shape: Sad face

For $f(x) = -4x^2 + 16$, we have $a=-4$, $b=0$, $c=16$.
The x-coordinate of the turning point is $x = -\frac{b}{2a} = -\frac{0}{2(-4)} = 0$.
The y-coordinate is $f(0) = -4(0)^2 + 16 = 16$.
Turning point: $\boxed{(0, 16)}$

Set $f(x) = 0$:

x-intercepts: $\boxed{(-2, 0) \text{ and } (2, 0)}$

The coefficient of $x^2$ is $a=-1$. Since $a < 0$, the parabola opens downwards.
Shape: Sad face

For $f(x) = -x^2 + 4$, we have $a=-1$, $b=0$, $c=4$.
The x-coordinate of the turning point is $x = -\frac{b}{2a} = -\frac{0}{2(-1)} = 0$.
The y-coordinate is $f(0) = -(0)^2 + 4 = 4$.
Turning point: $\boxed{(0, 4)}$

Set $f(x) = 0$:

x-intercepts: $\boxed{(-2, 0) \text{ and } (2, 0)}$

The coefficient of $x^2$ is $a=4$. Since $a > 0$, the parabola opens upwards.
Shape: Happy face

For $y = 4x^2 + 1$, we have $a=4$, $b=0$, $c=1$.
The x-coordinate of the turning point is $x = -\frac{b}{2a} = -\frac{0}{2(4)} = 0$.
The y-coordinate is $y = 4(0)^2 + 1 = 1$.
Turning point: $\boxed{(0, 1)}$

Set $y = 0$:

Since $x^2$ cannot be negative for real numbers, there are no real x-intercepts.
x-intercepts: $\boxed{\text{No real x-intercepts}}$