Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1$
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1.
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1.
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1.
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept: $\pm 2$ ; s-intercept: 1.
E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept: $\pm \sqrt { 8 }$ ;s-intercept: 2.

Question

Graph the quadratic function. Specify the vertex, axis of symmetry, maximum or minimum value, and intercepts. $s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1$
A) vertex: (- 2, 0); axis of symmetry: t = - 2; maximum value: 0; t-intercept: - 2; s-intercept: -1. 
B) vertex: (- 2, 0); axis of symmetry: t = - 2; minimum value: 0; t-intercept: - 2; s-intercept: 1. 
C) vertex: (2, 0); axis of symmetry: t = 2; maximum value: 0; t-intercept: 2; s-intercept: -1. 
D) vertex: (0,1); axis of symmetry: t = 0; maximum value: 1; t-intercept: $\pm 2$ ; s-intercept: 1. 
E) vertex: (0, 2);axis of symmetry: t = 0; maximum value: 2; t-intercept: $\pm \sqrt { 8 }$ ;s-intercept: 2.

Quizplus · Accepted Answer

We can determine the vertex of the quadratic function using the formula $(-\frac{b}{2a}, f(-\frac{b}{2a}))$. Plugging in the values, we get the vertex as $(-2,0)$. The coefficient of the squared term is negative, which means the parabola opens downwards and the vertex is a maximum point. Therefore, the maximum value is 0. The axis of symmetry is the line $t = -\frac{b}{2a}$, which here is $t = -2$. To find the $t$-intercepts, we set $s=0$ and solve the resulting quadratic equation $-\frac{1}{4}t^2-t-1=0$ to get $t=-2$ and $t=2$. To find the $s$-intercept, we set $t=0$ and solve for $s$ to get $s=-1$. Therefore, the correct choice is A.

Graph the Quadratic Function s=−14t2−t−1s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1s=−41​t2−t−1

Graph the Quadratic Function $s = - \frac { 1 } { 4 } t ^ { 2 } - t - 1$