Quadratic function (1 Viewer)

jnney · Aug 18, 2011

Y=4x-7 is a tangent to a parabola that has y int -3 and x=1/2 as axix of symmetry. Find the equation of the parabola

SpiralFlex · Aug 18, 2011

$y=ax^2+bx+c$ [That's our general quadratic.]

Passing through $(0, -3)$

$-3=a(0)^2+b(0)^2+c$

$c=-3$

Now, we got

$y=ax^2+bx-3$

We know that,

$-\frac{b}{2a}=\frac{1}{2}$ [Axis of symmetry/ $x$ coordinate of the vertex.]

$b=-a$

Hence,

$y=ax^2-ax-3$

Equate the line and curve.

$4x-7=a^2-ax-3$

$ax^2-(a+4)x+4=0$

$\Delta =(a+4)^2-16a$

For tangents, $\Delta = 0$

$\therefore (a+4)^2-16a=0$

$a^2+8a+16-16a=0$

$a^2-8a+16=0$

$(a-4)^2=0$

$a=4$

$\therefore$ The equation is $4x^2 - 4x -3 = 0$

Quadratic function (1 Viewer)

jnney

lemon

SpiralFlex

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