Answer to Question #114321 in Calculus for fawwaz hassan

Question

Determine the concavity of the parabola representing the quadratic function, its y intercept, its x
intercept if any exist, and the coordinates of the vertex. Sketch the parabola. Y = 25x
2 +8x-12.

Answer 1

Let us consider a parabola

$y(x) =ax^2+bx+c.$

If $a>0$, then for very large $x$ we get $y(x) >0,$ analogous for negative $x$with large modulus. Therefore, if $a>0$,then parabola is open (concave) up. In our example $a=25,$ so the parabola is concave up.

Now let us determine the vertex. The x-coordinate of vertex is

$x_0=\dfrac{-b}{2a}=\dfrac{-8}{2\cdot 25} = - 0.16.$

The second coordinate can be calculated as

$y_0=y(x_0)=25\cdot(-0.16)^2+8\cdot (-0.16)-12=0.64-1.28-12=-12.64.$

Next we obtain the intersection point of the parabola and y-axis. It is the point $(0, y(0))$ or $(0, - 12).$

Next we'll determine the intersection points of the parabola and x-axis. We shoul solve an equation

$25x^2+8x-12=0.$

$D=8^2+4\cdot25\cdot12 = 1264 =2^4\cdot79.$

Therefore, roots of equation are

$x_1 = - \dfrac{4+2\sqrt{79}} {25} ,\; x_2 = - \dfrac{4-2\sqrt{79}} {25}.$