Which form of this quadratic do i use to solve intercept and range?

Question

So my equation is: $-2x^2 + 4x + 30 = 0$

If I use this form to look at my y intercept, it will be 30. However, once I simplify it to: $x^2 - 2x - 15$, then my y intercept will be $-15$. Which one do I use? Inclusively, I was taught that if the quadratic is positive, it is in a U shape, and if it is negative, it is an upside-down U shape.

How do I know which form to use to solve these questions?

Answer 1

It does not simplify to $x^{2} - 2x - 15$, you have to continue to account for the $-2$ you factored out:

$$-2x^{2} + 4x + 30 = -2(x^{2} - 2x - 15) = 0 $$

Evaluating at $x = 0$, the intercept remains $(0,30)$.

Answer 2

Your equation is not $-2x^2 + 4x + 30 = 0$, but $f(x)=-2x^2 + 4x + 30 = -2(x^2-2x-15)$. You must find the $y$ intercept of a function, rather than an equation. To find the $y$ intercept, you need to let $x=0.$

$$f(0)=-2(0^2-2(0)-15)=-2(0-0-15)=-2(-15)=30$$

Thus, the $y$ intercept still occurs at $(0,30)$

When $-2x^2 + 4x + 30 = 0$, you are saying that you are trying to find where $f(x)=0$, and not a general function, $f(x)$. Note that $f(x)=-2x^2 + 4x + 30 \neq x^2-2x-15 $

Please comment if what I have said does not make sense.

Answer 3

The fact that you are interested in finding the range indicates that you were originally given the quadratic function $$f(x) = -2x^2 + 4x + 30$$ to graph. Its graph is the set of points $(x, y)$ in the Cartesian plane that satisfy the equation $y = -2x^2 + 4x + 30$.

Domain: The domain of any polynomial function is the set of all real numbers. Hence, the domain of $f(x) = -2x^2 + 4x + 30$ is $$D_f = \mathbb{R} = (-\infty, \infty)$$

Range: We complete the square to transform the equation from the standard form $$f(x) = ax^2 + bx + c$$ to the vertex form $$f(x) = a(x - h)^2 + k$$ where $(h, k)$ is the vertex of the graph of the parabola $y = ax^2 + bx + c$.
\begin{align*} f(x) & = -2x^2 + 4x + 30\\ & = -2(x^2 - 2x) + 30\\ & = -2\left[x^2 - 2x + \left(-\frac{2}{2}\right)^2\right] + 2\left(-\frac{2}{2}\right)^2 + 30\\ & = -2(x^2 - 2x + 1) + 2 + 30\\ & = -2(x - 1)^2 + 32 \end{align*} Hence, the graph of $f(x)$ has vertex $(1, 32)$. Since $(x - 1)^2 \geq 0$ for any real number $x$, $$-2(x - 1)^2 \leq 0 \implies -2(x - 1)^2 + 32 \leq 32$$
Thus, the function $f(x)$ has maximum value $32$, which is achieved when $x = 1$. Since $f(x) \leq 32$, the graph of the parabola opens downwards. Since $-2(x - 1)^2 + 32 \to -\infty$ as $x \to \infty$ or $x \to -\infty$, the range of $f(x)$ is $$R_f = (-\infty, 32] = \{x \in \mathbb{R} \mid x \leq -32\}$$ The graph opens downwards since the coefficient of $x^2$ is negative.

If you are not comfortable with completing the square, you can use the formula $$x = -\frac{b}{2a}$$ to find the equation of the axis of symmetry (the vertical line through the vertex of the parabola), then substitute that value of $x$ into the function to find the $y$-coordinate of the vertex.

Note: When the coefficient of $x^2$ is positive in the equation $$f(x) = ax^2 + bx + c = a(x - h)^2 + k$$ the graph opens upwards and has a minimum value of $k$ when $x = h$. On the graph, the minimum value occurs at the vertex $(h, k)$. The range of the function is $[k, \infty)$.

When the coefficient of $x^2$ is negative in the equation
$$f(x) = ax^2 + bx + c = a(x - h)^2 + k$$ the graph opens downwards and has a maximum value of $k$ at $x = h$. On the graph, the maximum value occurs at the vertex $(h, k)$. The range of the function is $(-\infty, k]$.

$\boldsymbol{x}$-intercepts: The $x$-intercepts occur where the graph intersects the $x$-axis, which is the line $y = 0$. To find the $x$-intercepts, we set $f(x) = 0$ and solve for $x$. \begin{align*} f(x) & = 0\\ -2x^2 + 4x + 30 & = 0 && \text{substitute $-2x^2 + 4x + 30$ for $f(x)$}\\ x^2 - 2x - 15 & = 0 && \text{divide each side of the equation by $2$}\\ (x - 5)(x + 3) & = 0 && \text{factor the resulting quadratic} \end{align*} Setting each factor equal to zero yields \begin{align*} x - 5 & = 0 & x + 3 & = 0\\ x & = 5 & x & = -3 \end{align*} which correspond to the points $(5, 0)$ and $(-3, 0)$ on the graph of the parabola $y = -2x^2 + 4x + 30$.

Alternatively, set $-2(x - 1)^2 + 32 = 0$ and solve for $x$. \begin{align*} f(x) & = 0\\ -2(x - 1)^2 + 32 & = 0 && \text{substitute $-2(x - 1)^2 + 32$ for $f(x)$}\\ (x - 1)^2 - 16 & = 0 && \text{divide each side of the equation by $-2$}\\ (x - 1)^2 & = 16 && \text{add $16$ to each side of the equation}\\ |x - 1| & = 4 && \text{take square roots} \end{align*} Since $|x - 1| = 4$ means the number $x - 1$ is $4$ units from zero, either $x - 1 = 4$ or $x - 1 = -4$.
\begin{align*} x - 1 & = 4 & x - 1 & = -4\\ x & = 5 & x & = -3 \end{align*}

$\boldsymbol{y}$-intercept: The $y$-intercept occurs where the graph of the function intersects the $y$-axis, which is the line $x = 0$. To solve for the $y$-intercept, evaluate $f(x)$ when $x = 0$. \begin{align*} f(x) & = -2x^2 + 4x + 30\\ f(0) & = -2 \cdot 0^2 + 4 \cdot 0 + 30 && \text{substitute $0$ for $x$}\\ & = -2 \cdot 0 + 0 + 30\\ & = 0 + 0 + 30\\ & = 30 \end{align*} On the graph of the parabola $y = -2x^2 + 4x + 30$, the $y$-intercept is the point $(0, 30)$.