Parabolas: Additional Practice (Set 2)

Compute the discriminant with $a = 1$, $b = 2$, $c = 5$.

$D = (2)^2 - 4(1)(5) = 4 - 20 = -16$

Since $D < 0$ there are no real solutions, so the parabola never meets the x-axis and has no x-intercepts.

The vertex x-coordinate is $-\tfrac{b}{2a}$ with $a = 1$, $b = -10$.

$x = -\frac{-10}{2} = 5$

Substitute: $y = (5)^2 - 10(5) + 21 = 25 - 50 + 21 = -4$. The vertex is $(5, -4)$.

Mirror images across the axis of symmetry share a height. The point $x = 2$ is $3$ units right of the axis $x = -1$, so its twin is $3$ units left, at the same $y = 5$.

$-1 - (2 - (-1)) = -1 - 3 = -4 \;\Rightarrow\; (-4, 5)$

So $(-4, 5)$ must also be on the parabola.

Use $x = -\tfrac{b}{2a}$ with $a = 4$, $b = -4$.

$x = -\frac{-4}{2(4)} = \frac{4}{8} = \frac{1}{2}$

So the axis of symmetry is $x = \tfrac{1}{2}$.

The point $(0, 5)$ is the y-intercept, and setting $x = 0$ leaves $y = c$.

$y = (0)^2 + b(0) + c = c = 5$

So $c = 5$, regardless of the value of $b$.

The parabola opens upward, so the minimum is the vertex height. The vertex x-coordinate is $-\tfrac{b}{2a} = -\tfrac{8}{2} = -4$.

$y = (-4)^2 + 8(-4) + 20 = 16 - 32 + 20 = 4$

So the smallest value of $y$ is $4$.

An upward parabola has its vertex as the lowest point. If that lowest point is already above the x-axis, the whole curve stays above it.

$\text{lowest point above the axis} \;\Rightarrow\; \text{never reaches } y = 0$

So the parabola has no x-intercepts.

No x-intercepts means the discriminant is negative, with $a = 1$, $b = 6$.

$D = 6^2 - 4(1)(c) = 36 - 4c < 0 \;\Rightarrow\; c > 9$

Among the choices only $c = 10$ satisfies $c > 9$. At $c = 9$ the discriminant is $0$ (one intercept), and $c = 0$ or $5$ give two.

A downward parabola has its vertex as the highest point. If that highest point is already below the x-axis, the entire curve stays below it.

$\text{highest point below the axis} \;\Rightarrow\; \text{never reaches } y = 0$

So the parabola has no x-intercepts.

The y-intercept is at $x = 0$, where the graph meets the y-axis. Setting $x = 0$ leaves the constant term.

$y = (0)^2 - 4(0) + 3 = 3$

So the graph crosses the y-axis at $(0, 3)$.

Use $x = -\tfrac{b}{2a}$ with $a = 1$, $b = -7$.

$x = -\frac{-7}{2(1)} = \frac{7}{2}$

So the axis of symmetry is $x = \tfrac{7}{2}$.

An upward parabola has its vertex as the lowest point, so the minimum value of $y$ is the vertex's y-coordinate.

$\text{vertex } (3, -4) \;\Rightarrow\; \text{minimum } y = -4$

Every other point on the curve is higher than $-4$.