Graphing Quadratic Equations: exemplo 10.48

Problema e solução em inglês, como no livro original.

Find the intercepts of the parabola $y = 4 x^{2} - 12 x + 9$ .

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Sua vez

O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.

Find the intercepts of the parabola $y = − x^{2} - 12 x - 36 .$

Como foi?


To find the y-intercept, let $x = 0$ and solve for y.
		When $x = 0$ , then $y = 9$ . The y-intercept is the point $(0, 9)$ .

To find the x-intercept, let $y = 0$ and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.		$\begin{matrix} \begin{matrix} b^{2} & - & 4 a c \\ 2^{2} & - & 4 \cdot 4 \cdot 9 \\ 144 & - & 144 \end{matrix} \\ 0 \end{matrix}$
		Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x-intercept.
Solve the equation by factoring the perfect square trinomial.
Use the Zero Product Property.
Solve for x.
		When $y = 0$ , then $\frac{3}{2} = x .$
		The x-intercept is the point $(\frac{3}{2}, 0) .$


To find the y-intercept, let $x = 0$ and solve for y.
		When $x = 0$ , then $y = 9$ . The y-intercept is the point $(0, 9)$ .

To find the x-intercept, let $y = 0$ and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.		$\begin{matrix} \begin{matrix} b^{2} & - & 4 a c \\ 2^{2} & - & 4 \cdot 4 \cdot 9 \\ 144 & - & 144 \end{matrix} \\ 0 \end{matrix}$
		Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x-intercept.
Solve the equation by factoring the perfect square trinomial.
Use the Zero Product Property.
Solve for x.
		When $y = 0$ , then $\frac{3}{2} = x .$
		The x-intercept is the point $(\frac{3}{2}, 0) .$