25 ++ y=x^2-4x graph 531403-Y=x^2+4x+7 graph

As a is positive, the parabola should open up As c=0, the parabola passes through the origin Vertex x=−b/2a=2, so y=−4 ⇒Vertex=(2,−4) Assuming two values of x, x 1 0 y

Y=x^2+4x+7 graph- Set y=0 to find the x intercept(s) if they exist #y = x^2 4x 1# #0 = x^2 4x 1# use the quadratic formula #x = (bsqrt(b^2 4ac))/(2a)# #x = (4sqrt(4^2 4*1*1))/(2*1)# #x=2sqrt3# Now find the Axis of Symmetry (aos) #aos = b/(2a) = (4)/(2*1) = 2# Find the vertex vertex = (aos, f(aos)) #= (2, f(2))# #f(x) = x^2 4x 1# #f(2) = (2)^2 4* 2 1 = 3#The minimum value of the quadratic is −9 − 9 and it occurs when x = −1 x = − 1 Show the graph to verify the result We have used the formula h = −16t2 v0t h0 h = − 16 t 2 v 0 t h 0 to calculate the height in feet, h h, of an object shot upwards into

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How do I represent the graph of the parabola y^2=4x? STEP 1 Find the line of symmetry Example y = 2x2 4x 1 ( ) 4 1 2 2 2 b x a = = = y x Thus theHow To Graph A Quadratic Equation 10 Steps With Pictures y = x2 y = ‑ x2 Wider than parent function, reflected in xaxis Given a parabola in Standard Form, y = ax2 bx c, • When a > 0, the parabola opens up • When a 0, the parabola

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