Home > Algebra > Quadratic Functions and Their Graphs

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y=x^2-7x-28#?

Answer 1

Find vertex of y = x^2 - 7x - 28

x of vertex: #x = (-b/2a) = 7/2#

y of vertex: #y = f(7/2) = 49/4 - 49/2 - 28 = 161/4#

x of axis of symmetry = x of vertex =# 7/2#

To find y-intercept, make x = 0 -> y = -28 To find x-intercepts, make y = 0 #D = d^2 = b^2 - 4ac# = 49 + 112 = 161 --># d = +- sqrt161#

x-intercepts:# x = 7/2 +- (sqrt161)/2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Nathan Axel

To find the vertex of a quadratic equation in the form y = ax^2 + bx + c, use the formula x = -b/(2a). Then substitute the x-value into the equation to find the corresponding y-value. The vertex is (x, y).

The axis of symmetry is the vertical line that passes through the vertex. Its equation is x = -b/(2a).

To find the y-intercept, set x = 0 in the equation and solve for y. To find the x-intercepts, set y = 0 in the equation and solve for x using the quadratic formula or factoring.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.