Parabolas: Practice

Use the discriminant with $a = 2$, $b = -3$, $c = -4$, watching the double negative.

$D = (-3)^2 - 4(2)(-4) = 9 + 32 = 41$

Since $D > 0$ there are two distinct real x-intercepts.

The vertex x-coordinate is $x = -\tfrac{b}{2a} = -\tfrac{5}{2}$. Substitute it back for the height.

$y = \left(-\tfrac{5}{2}\right)^2 + 5\left(-\tfrac{5}{2}\right) + 6 = \tfrac{25}{4} - \tfrac{50}{4} + \tfrac{24}{4} = -\tfrac{1}{4}$

So the vertex is $\left(-\tfrac{5}{2}, -\tfrac{1}{4}\right)$.

Vertex form $y = a(x - h)^2 + k$ needs $(h, k) = (3, -2)$, so $x - 3$ inside and $-2$ outside, with $a > 0$ for upward.

$y = (x - 3)^2 - 2$

The others fail: $(x + 3)^2 - 2$ has vertex $(-3, -2)$; $-(x - 3)^2 - 2$ opens downward; $(x - 3)^2 + 2$ has vertex $(3, 2)$.

The leading coefficient $a = -2$ is negative, so the parabola opens downward and its vertex is the highest point. In vertex form the vertex height is $k$.

$y = -2(x + 1)^2 + 8 \;\Rightarrow\; k = 8$

The squared term only subtracts from $8$, so the maximum value of $y$ is $8$, reached at $x = -1$.

Exactly one x-intercept means the discriminant is zero, with $a = 1$, $b = -4$, $c = k$.

$(-4)^2 - 4(1)(k) = 0 \;\Rightarrow\; 16 - 4k = 0 \;\Rightarrow\; k = 4$

At $k = 4$ the quadratic is $x^2 - 4x + 4 = (x - 2)^2$, touching the axis once at $x = 2$.

The parabola opens upward ($a = 3 > 0$) with vertex $(0, 1)$, which is above the x-axis, so the whole curve floats above it. The discriminant confirms this with $b = 0$, $c = 1$.

$D = 0^2 - 4(3)(1) = -12 < 0$

A negative discriminant means no real x-intercepts, so the parabola never crosses the x-axis.

Factor $2$ from the first two terms, complete the square inside, and simplify.

$y = 2(x^2 + 2x) + 5 = 2\left((x + 1)^2 - 1\right) + 5 = 2(x + 1)^2 + 3$

In vertex form this is $y = 2(x + 1)^2 + 3$, so the vertex is $(-1, 3)$.

One x-intercept means the curve touches the axis at a single point, which happens when the quadratic is a perfect square (discriminant $0$).

$y = (x - 2)^2 = 0 \;\Rightarrow\; x = 2 \text{ only}$

The others differ: $x^2 + 4$ has none, while $x^2 - 4$ and $x^2 - x - 2 = (x - 2)(x + 1)$ each have two.

Complete the square: half of $-4$ is $-2$, and $(-2)^2 = 4$.

$y = (x^2 - 4x + 4) - 4 + 7 = (x - 2)^2 + 3$

So $y = (x - 2)^2 + 3$, and the vertex is $(2, 3)$.

The y-intercept is at $x = 0$, so substitute $x = 0$ into the vertex form and simplify.

$y = (0 - 3)^2 + 1 = 9 + 1 = 10$

So the y-intercept is $(0, 10)$. The constant $1$ is the vertex height, not the y-intercept.

The parabola opens upward, so $y$ is smallest at the vertex, whose x-coordinate is $-\tfrac{b}{2a}$ with $a = 1$, $b = 6$.

$x = -\frac{6}{2(1)} = -3$

So $y$ reaches its minimum at $x = -3$.

Compute the discriminant with $a = -1$, $b = 4$, $c = -4$.

$D = (4)^2 - 4(-1)(-4) = 16 - 16 = 0$

Since $D = 0$ there is exactly one x-intercept. Indeed $-x^2 + 4x - 4 = -(x - 2)^2$, touching the axis at $x = 2$.